Derivatives of Trigonometric Functions
Greek letters being usual to denote angles, we will take as the usual letter for any variable angle the letter \(\theta\) (“theta”). In this chapter, \(\theta\) is measured in radians.1
Derivative of Sine
Let us consider the function \[y= \sin \theta.\]
What we have to investigate is the value of \(\dfrac{d(\sin \theta)}{d \theta}\); or, in other words, if the angle \(\theta\) varies, we have to find the relation between the increment of the sine and the increment of the angle, both increments being indefinitely small in themselves. Examine the next figure, wherein, if the radius of the circle is unity, the height of \(y\) is the sine, and \(\theta\) is the angle. Now, if \(\theta\) is supposed to increase by the addition to it of the small angle \(d \theta\)—an element of angle—the height of \(y\), the sine, will be increased by a small element \(dy\). The new height \(y + dy\) will be the sine of the new angle \(\theta + d \theta\), or, stating it as an equation, \[y+dy = \sin(\theta + d \theta);\] and subtracting from this the first equation gives \[dy = \sin(\theta + d \theta)- \sin \theta.\]
The quantity on the right-hand side is the difference between two sines, and books on trigonometry tell us how to work this out. For they tell us that if \(M\) and \(N\) are two different angles, \[\sin M - \sin N = 2 \cos\frac{M+N}{2}\cdot\sin\frac{M-N}{2}.\]
If, then, we put \(M= \theta + d \theta\) for one angle, and \(N= \theta\) for the other, we may write \[\begin{align} dy &= 2 \cos\frac{\theta + d\theta + \theta}{2} \cdot \sin\frac{\theta + d\theta - \theta}{2}, \end{align}\] or, \[\begin{align} dy &= 2\cos\left(\theta + \frac{1}{2}d\theta\right)\cdot \sin\frac{d\theta}{2}. \end{align}\]
But if we regard \(d \theta\) as indefinitely small, then in the limit we may neglect \(\frac{1}{2} d \theta\) by comparison with \(\theta\), and may also take \(\sin\frac{d\theta}{2}\) as being the same as \(\frac{1}{2} d \theta\). The equation then becomes: \[\begin{align} dy &= 2 \cos \theta \times \frac{1}{2} d \theta; \\ dy &= \cos \theta \cdot d \theta, \end{align}\] and, finally, \[\begin{align} \dfrac{dy}{d \theta} &= \cos \theta. \end{align}\] [Notice that the approximation \(\sin \frac{d\theta}{2}\approx \frac{d\theta}{2}\) is true only when \(d\theta\) is measured in radians.]
The accompanying curves in the next two figures show, plotted to scale, the values of \(y=\sin \theta\), and \(\dfrac{dy}{d\theta}=\cos\theta\), for the corresponding values of \(\theta\).
Derivative of Cosine
Take next the cosine.
Let \(y=\cos \theta\).
Now \(\cos \theta=\sin\left(\dfrac{\pi}{2}-\theta\right)\).
Therefore \[\begin{align} dy = d\left(\sin\left(\frac{\pi}{2} - \theta\right)\right) &= \cos\left(\frac{\pi}{2} - \theta\right) \times d(-\theta), \\ &= \cos\left(\frac{\pi}{2} - \theta\right) \times (-d\theta), \end{align}\] \[\frac{dy}{d\theta} = -\cos\left(\frac{\pi}{2} - \theta\right).\] And it follows that \[\frac{dy}{d\theta} = -\sin \theta.\]
Derivative of Tangent
Lastly, take the tangent.
As \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\), we can apply the Quotient Rule to find \(\dfrac{d(\tan\theta)}{d\theta}\):2
\[\begin{align} \frac{d(\tan\theta)}{d\theta}&=\frac{\cos\theta \dfrac{d(\sin\theta)}{d\theta}-\sin\theta \dfrac{d(\cos\theta)}{d\theta}}{\cos^2\theta}\\ &=\frac{\cos\theta \cdot \cos\theta -\sin\theta (-\sin\theta)}{\cos^2\theta}\\ &=\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta} \end{align}\] Since \[\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}=1+\left(\frac{\sin \theta}{\cos \theta}\right)^2\] we get \[\frac{d(\tan\theta)}{d\theta}=1+\tan^2 \theta.\] Also, since \(\sin^2\theta+\cos^2\theta=1\), we get \[\frac{d(\tan\theta)}{d\theta}=\frac{1}{\cos^2\theta}=\sec^2\theta.\] Therefore, \[\frac{d(\tan\theta)}{d\theta}=\sec^2\theta=1+\tan^2\theta.\]
Summary of Results
Collecting these results, we have: \[\begin{array}{|c|c|} \hline y & \dfrac{dy}{d\theta} \\[9pt] \hline \sin\theta & \cos\theta \\[9pt] \cos\theta & -\sin\theta\\[9pt] \tan\theta & \sec^2\theta=1+\tan^2 \theta\\[9pt] \hline \end{array}\]
To derive the above results, we replaced \(\sin(d\theta/2)\) by \(d\theta/2\). In general, \(\sin x\) is approximately equal to \(x\) when (1) \(x\) is small (2) \(x\) is measured in radians \[\sin x\approx x \qquad (x \text{ is small and measured in radians}).\] For example, \(1^\circ\) is the same as \(\frac{\pi}{180}\) radians, and we cannot approximate \(\sin 1^\circ\) by 1 \[\sin 1^\circ\neq 1\] but \[\sin 1^\circ=\sin\frac{\pi}{180}\approx \frac{\pi}{180}=0.017.\] Therefore, the results tabulated above are true only when \(\theta\) is measured in radians.
Sometimes, in mechanical and physical questions, as, for example, in simple harmonic motion and in wave-motions, we have to deal with angles that increase in proportion to the time. Thus, if \(T\) be the time of one complete period, or movement round the circle, then, since the angle all round the circle is \(2\pi\) radians, (equivalent to \(360^\circ\)), the amount of angle moved through in time \(t\), will be \[\begin{align} \theta &= 2\pi\frac{t}{T},\quad \text{in radians.} \end{align}\] If the frequency, or number of periods per second, be denoted by \(n\), then \(n = \dfrac{1}{T}\), and we may then write: \[\theta=2\pi nt.\] Then we shall have \[y = \sin (2\pi nt).\]
If, now, we wish to know how the sine varies with respect to time, we must differentiate with respect, not to \(\theta\), but to \(t\). For this we must resort to the chain rule explained in the chapter on the Chain Rule, and put \[\frac{dy}{dt} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dt}.\]
Now \(\dfrac{d\theta}{dt}\) will obviously be \(2\pi n\); so that \[\begin{align} \frac{dy}{dt} &= \cos \theta \times 2\pi n \\ &= 2\pi n \cdot \cos (2\pi nt). \end{align}\] Similarly, it follows that \[\begin{align} \frac{d\left(\cos (2\pi nt)\right)}{dt} &= -2\pi n \cdot \sin (2\pi nt). \end{align}\]
Second Derivatives of Sine and Cosine
We have seen that when \(\sin \theta\) is differentiated with respect to \(\theta\) it becomes \(\cos \theta\); and that when \(\cos \theta\) is differentiated with respect to \(\theta\) it becomes \(-\sin \theta\); or, in symbols, \[\frac{d^2({\sin \theta})}{d\theta^2} = -\sin \theta.\]
So we have this curious result that we have found a function such that if we differentiate it twice over, we get the same thing from which we started, but with the sign changed from \(+\) to \(-\).
The same thing is true for the cosine; for differentiating \(\cos\theta\) gives us \(-\sin\theta\), and differentiating \(-\sin\theta\) gives us \(-\cos\theta\); or thus: \[\frac{d^2(\cos\theta)}{d\theta^2} = -\cos\theta.\]
Sines and cosines are the only functions of which the second derivative is equal (and of opposite sign to) the original function.
Examples
With what we have so far learned we can now differentiate expressions of a more complex nature.
Example 15.1. If \(y=\arcsin x\), find \(\dfrac{dy}{dx}\).
[In many modern calculus textbooks, the inverse sine is denoted by \(\sin^{-1}\); that is, \(\arcsin x=\sin^{-1} x\). Note that \(\sin^{-1} x\) is NOT the same as \(\frac{1}{\sin x}\). To avoid confusion, the notation \(\arcsin x\) may be preferred over \(\sin^{-1} x\) in some texts, including this one.]
Solution. If \(y\) is the arc whose sine is \(x\), then \(x = \sin y\). \[\frac{dx}{dy}=\cos y.\]
Passing now from the inverse function to the original one, we get \[\begin{align} \frac{dy}{dx} &= \frac{1}{\;\dfrac{dx}{dy}\;} = \frac{1}{\cos y}. \end{align}\] now since \(\cos^2 y+\sin^2 y=1\), \[\cos y= \pm \sqrt{1-\sin^2 y}=\pm \sqrt{1-x^2};\] hence \[\begin{align} \frac{dy}{dx} &=\frac{1}{\cos y}= \pm \frac{1}{\sqrt{1-x^2}}. \end{align}\] But which one is correct? plus or minus? If we look at the graph of \(y=\arcsin x\) (the following figure), we realize that the slope of the curve is always positive, indicating that that we must take the positive square root. Hence, \[\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}.\]
\[\begin{align} \boxed{\dfrac{d(\arcsin x)}{dx}=\dfrac{1}{\sqrt{1-x^2}}} \end{align}\]
Example 15.2. If \(y=\cos^3 \theta\), find \(\dfrac{dy}{d\theta}\).
Solution. This is the same thing as \(y=(\cos \theta)^3\).
Let \(\cos\theta=v\);then \(y=v^3\);\(\dfrac{dy}{dv}=3v^2\). \[\begin{align} \frac{dv}{d\theta} &= -\sin\theta.\\ \frac{dy}{d\theta} &= \frac{dy}{dv} \times \frac{dv}{d\theta} = -3 \cos^2 \theta \sin\theta. \end{align}\]
Example 15.3. If \(y=\sin(x+a)\), find \(\dfrac{dy}{dx}\).
Solution. Let \(x+a=v\);then \(y=\sin v\). \[\frac{dy}{dv}=\cos v;\qquad \frac{dv}{dx}=1 \quad\text{and}\quad \frac{dy}{dx}=\cos(x+a).\]
Example 15.4. If \(y=\ln \sin \theta\), find \(\dfrac{dy}{d\theta}\).
Solution. Let \(\sin\theta=v\);\(y=\ln v\). \[\begin{align} \frac{dy}{dv} &= \frac{1}{v};\quad \frac{dv}{d\theta}=\cos\theta;\\ \frac{dy}{d\theta} &= \frac{1}{\sin\theta} \times \cos\theta = \cot\theta. \end{align}\]
Example 15.5. If \(y=\cot\theta=\dfrac{\cos\theta}{\sin\theta}\), find \(\dfrac{dy}{d\theta}\).
Solution. \[\begin{align} \frac{dy}{d\theta} &= \frac{-\sin^2\theta - \cos^2 \theta}{\sin^2 \theta}\\ &= -(1+\cot^2 \theta) = -\text{csc}^2 \theta. \end{align}\]
Example 15.6. If \(y=\tan 3\theta\), find \(\dfrac{dy}{d\theta}\).
Solution. Let \(3\theta=v\);\(y=\tan v\);\(\dfrac{dy}{dv}=\sec^2 v\). \[\frac{dv}{d\theta}=3;\quad \frac{dy}{d\theta}=3 \sec^2 3\theta.\]
Example 15.7. If \(y = \sqrt{1+3\tan^2\theta}\), find \(\dfrac{dy}{d\theta}\).
Solution. \(y=(1+3 \tan^2 \theta)^{\frac{1}{2}}\).
Let \(3\tan^2\theta=v\). \[y = (1+v)^{\frac{1}{2}};\quad \frac{dy}{dv} = \frac{1}{2\sqrt{1+v}} \] \[\frac{dv}{d\theta} = 6\tan\theta \sec^2 \theta \] (for, if \(\tan \theta = u\), \[\begin{align} v &= 3u^2;\quad \frac{dv}{du} = 6u;\quad \frac{du}{d\theta} = \sec^2 \theta; \end{align}\] hence \(\displaystyle \frac{dv}{d\theta}= 6 \tan \theta \sec^2 \theta\);)
hence \[\begin{align} \frac{dy}{d\theta} &= \frac{6\tan\theta \sec^2\theta}{2\sqrt{1 + 3\tan^2\theta}}. \end{align}\]
Example 15.8. If \(y=\sin x \cos x\), find \(\dfrac{dy}{dx}\) .
Solution. \[\begin{align} \frac{dy}{dx} &= \sin x(-\sin x) + \cos x \times \cos x = \cos^2 x - \sin^2 x. \end{align}\]
Exercises
Exercise 15.1. Differentiate the following: \[\begin{align} \text{(i)}\quad y &= A \sin\left(\theta - \frac{\pi}{2}\right).\\ \text{(ii)}\quad y &= \sin^2 \theta;\quad \text{and }\quad y = \sin 2\theta.\\ \text{(iii)}\quad y &= \sin^3 \theta;\quad \text{and }\quad y = \sin 3\theta. \end{align}\]
Answer
(i) \(\dfrac{dy}{d\theta} = A \cos \left( \theta - \dfrac{\pi}{2} \right)\);
(ii) \(\dfrac{dy}{d\theta} = 2\sin\theta \cos\theta = \sin2\theta\) and \(\dfrac{dy}{d\theta} = 2\cos2\theta\);
(iii) \(\dfrac{dy}{d\theta} = 3\sin^2 \theta \cos\theta\) and \(\dfrac{dy}{d\theta} = 3\cos3\theta\).
Solution
(i) \(\displaystyle y=A \sin \left(\theta-\frac{\pi}{2}\right)\)
We write \[y=A \sin u \quad \text { where }\quad u=\theta-\frac{\pi}{2}\] Using the Chain Rule: \[\begin{align} \frac{d y}{d \theta}&= \frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ & =(A \cos u)(1) \\ & =A \cos \left(\theta-\frac{\pi}{2}\right) \end{align}\]
(ii) If \(y =\sin ^{2} \theta=(\sin \theta)^{2}\)
Let \[y =u^{2} \quad \text { where }\quad u=\sin \theta\] Using the Chain Rule: \[\begin{align} \frac{d y}{d \theta} & =\frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ & =2 u \cdot \cos \theta \\ & =2 \sin \theta \cos \theta \end{align}\] The result can also be written as \(\sin 2 \theta\) since \(\sin 2 \theta=2 \sin \theta \cos \theta.\)
If \(y=\sin 2 \theta\), let \[y=\sin u\quad \text { where }\quad u=2 \theta.\] Using the Chain Rule: \[\begin{align} \frac{d y}{d \theta}&=\frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ &=(\cos u)(2) \\ &=2 \cos (2 \theta) \end{align}\] If \(y=\sin ^{3} \theta=(\sin \theta)^{3}\), we write \[y=u^{3}\quad \text{where}\quad u=\sin \theta\] Then using the Chain Rule \[\begin{align} \frac{d y}{d \theta}&=\frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ &=3 u^{2} \cdot \cos \theta \\ &=3(\sin \theta)^{2} \cdot \cos \theta \\ &=3 \cos \theta \sin ^{2} \theta \\ &=(\cos u) \cdot(3) \\ &=3 \cos 3 \theta \cdot \sin 3 \theta \end{align}\] If \(y=\sin 3\theta\), we write \[y=\sin u\quad\text{where}\quad u=3\theta\] Then \[\begin{align} \frac{dy}{d \theta}&=\frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ &=(\cos u)(3)\\ &=3\cos {3\theta} \end{align}\]
Exercise 15.2. Find the value of \(\theta\) for which \(\sin\theta \times \cos\theta\) is a maximum.
Answer
\(\theta = 45^\circ\) or \(\dfrac{\pi}{4}\) radians.
Solution
\[y=\sin \theta \cos \theta\]
Method 1) Using the Product Rule:
\[\begin{align} \frac{d y}{d \theta}&=\cos \theta \cdot \cos \theta-\sin \theta \sin \theta \\ &=\cos ^{2} \theta-\sin ^{2} \theta \\ &=\cos 2 \theta \end{align}\] \[\frac{d y}{d \theta}=\cos 2 \theta=0\ \Leftrightarrow\ 2 \theta=\frac{\pi}{2}\ \text { or }\ 2 \theta=\frac{3 \pi}{2}\] \[\frac{d y}{d \theta}=0\quad \Leftrightarrow \quad \theta=\frac{\pi}{4} \quad \text { or } \quad \theta=\frac{3 \pi}{4}\] \[\begin{align} \frac{d^{2} y}{d \theta^{2}}&= \frac{d(\cos 2 \theta)}{d(2 \theta)} \cdot \frac{d(2 \theta)}{d \theta} \\ &=(-\sin 2 \theta)(2) \\ &=-2 \sin 2 \theta \end{align}\] When \(\theta=\dfrac{\pi}{4}\) \[\frac{d^{2} y}{d \theta^{2}}=-2<0\] Hence, the curve is concave downward, and thus when \(\theta=\dfrac{\pi}{4}\), \(y\) has a maximum of \[\sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{1}{2}.\]
When \(\theta=\dfrac{3 \pi}{4}\) \[\frac{d^{2} y}{d \theta^{2}}=-2 \sin \left(\frac{3 \pi}{2}\right)=2>0\] hence, the curve is concave upward and thus when \(\theta=\dfrac{3 \pi}{4}\), \(y\) has a minimum of \[\begin{align} \sin \left(\frac{3 \pi}{4}\right) \cos \left(\frac{3 \pi}{4}\right) & =\sin \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \cos \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \\ & =\cos \left(\frac{\pi}{4}\right) \times\left(-\sin \left(\frac{\pi}{4}\right)\right) \\ & =\frac{1}{\sqrt{2}} \times \frac{-1}{\sqrt{2}} \\ & =-\frac{1}{2}. \end{align}\]
Method 2)
\[y=\sin \theta \cdot \cos \theta=\frac{1}{2} \sin 2 \theta\] \(y\) is a maximum wherever \(\sin 2 \theta\) is a maximum and that occurs when \[2 \theta=\frac{\pi}{2} \text { or } \quad \theta=\frac{\pi}{4}\] The maximum of \(y\) is then \(\frac{1}{2} \cdot \sin \left(\frac{\pi}{2}\right)=\frac{1}{2}\).
Exercise 15.3. Differentiate \(y=\dfrac{1}{2\pi} \cos (2\pi nt)\).
Answer
\(\dfrac{dy}{dt} = -n \sin 2\pi nt\).
Solution
\[y=\frac{1}{2 \pi} \cos (2 \pi n t)\] We write \[y=\frac{1}{2 \pi} \cos u \quad \text { where }\quad u=2 \pi n t\] \[\begin{align} \frac{d y}{d t} &= \frac{d y}{d u} \cdot \frac{d u}{d t} \\ &=-\frac{1}{2 \pi} \sin u \cdot 2 \pi n \\ &=-n \sin u \\ &= n \sin (2 \pi n t) \end{align}\]
Exercise 15.4. If \(y = \sin a^x\), find \(\dfrac{dy}{dx}\).
Answer
\(a^x \ln a \cos a^x\).
Solution
\[y=\sin \left(a^{x}\right)\] Let \(y=\sin u\) where \(u=a^{x}\). Then \[\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\cos u \cdot a^{x} \cdot \ln a \\ & =\cos \left(a^{x}\right) \cdot a^{x} \cdot \ln a \end{align}\]
Exercise 15.5. Differentiate \(y=\ln \cos x\).
Answer
\(\dfrac{\cos x}{\sin x} = \cot x\)
Solution
\[y=\ln \cos x\]
We write \(y=\ln u\) where \(u=\cos x\). Then
\[\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\frac{1}{u} \cdot(-\sin x) \\ & =\frac{1}{\cos x}(-\sin x) \\ & =-\tan x \end{align}\]
Exercise 15.6. Differentiate \(y=18.2 \sin(x+26)\).
Answer
\(18.2 \cos \left(x + 26 \right)\).
Solution
\[y=18.2 \sin (x+26)\]
We wrtie \(y=18.2 \sin u\) where \(u=x+26\). Then
\[\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =(18.2 \cos u)(1) \\ & =18.2 \cos (x+26) \end{align}\]
Exercise 15.7. Plot the curve \(y=10 \sin\left(\theta-\dfrac{\pi}{12}\right)\); and show that the slope of the curve at \(\theta = \dfrac{5\pi}{12}\) is half the maximum slope.
Answer
The slope is \(\dfrac{dy}{d\theta} = 100\cos\left(\theta -\frac{\pi}{12} \right)\), which is a maximum when \((\theta -\frac{\pi}{12}) = 0\), or \(\theta = \frac{\pi}{12}\); the value of the slope being then \({}= 100\). When \(\theta = \frac{5\pi}{12}\) the slope is \(100\cos\left(\frac{5\pi}{12} - \frac{\pi}{12}\right) = 100\cos \frac{\pi}{3} = 100 \times \frac{1}{2} = 50\).
Solution
\[\begin{align} & y=10 \sin \left(\theta-\frac{\pi}{12}\right) \\ & \frac{d y}{d \theta}=10 \cos \left(\theta-\frac{\pi}{12}\right) \end{align}\]
To find the maximum slope, we have to differentiate \(\dfrac{d y}{d \theta}\) with respect to \(\theta\) and equate the result to zero
\[\frac{d\left(\dfrac{d y}{d \theta}\right)}{d \theta}=\frac{d^{2} y}{d \theta^{2}}=-10 \sin \left(\theta-\frac{\pi}{12}\right)=0\]
\[\begin{align} \frac{d^{2} y}{d \theta^{2}}=0\quad\Leftrightarrow &\quad \theta-\frac{\pi}{12}=0 \quad \text { or } \quad \theta-\frac{\pi}{12}=\pi \\ \frac{d^{2} y}{d \theta^{2}}=0\quad \Leftrightarrow &\quad \theta=\frac{\pi}{12} \quad \text { or } \quad \theta=\frac{13 \pi}{12} \end{align}\]
When \(\theta=\dfrac{\pi}{12}\) \[\frac{d y}{d \theta}=10 \cos 0=10\quad (\text {max slope})\]
When \(\theta=\dfrac{13 \pi}{12}\) \[\frac{d y}{d \theta}=10 \cos (\pi)=-10 \quad (\text {min slope})\]
Slope when \(\theta=\dfrac{5 \pi}{12}\): \[\begin{align} \frac{d y}{d \theta} & =10 \cos \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)=10 \cos \left(\frac{\pi}{3}\right) \\ & =10 \times \frac{1}{2}=5 . \end{align}\] As we can see the slope of the curve at \(\theta=\frac{5 \pi}{12}\), which is 5, is half the maximum slope, which is 10 and occurs when \(\theta=\frac{\pi}{12}.\)
Exercise 15.8. If \(y=\sin \theta\cdot\sin 2\theta\), find \(\dfrac{dy}{d\theta}\).
Answer
\[\begin{align} \cos\theta \sin2\theta + 2\cos2\theta \sin\theta &= 2\sin\theta\left(\cos^2 \theta + \cos2\theta\right) \\ &= 2\sin\theta\left(3\cos^2 \theta - 1\right). \end{align}\]
Solution
\[y=\sin \theta \cdot \sin 2 \theta\]
Using the Product Rule:
\[\frac{d y}{d \theta}=\frac{d(\sin \theta)}{d \theta} \cdot \sin 2 \theta+\sin \theta \frac{d(\sin 2 \theta)}{d \theta}\]
We showed in Exercise 1 (ii) \(\dfrac{d(\sin 2 \theta)}{d \theta}=2 \cos 2 \theta\).
Hence
\[\frac{d y}{d \theta}=\cos \theta \cdot \sin 2 \theta+2 \sin \theta \cos 2 \theta\]
We can further simplify it using
\[\sin 2 \theta=2 \sin \theta \cos \theta\] and \[\cos 2 \theta=2 \cos ^{2} \theta-1.\]
\[\begin{align} \frac{d y}{d \theta} & =2 \sin \theta \cos ^{2} \theta+2 \sin \theta\left(2 \cos ^{2} \theta-1\right) \\ & =2 \sin \theta\left(\cos ^{2} \theta+2 \cos ^{2} \theta-1\right) \\ & =2 \sin \theta\left(3 \cos ^{2} \theta-1\right) . \end{align}\]
Exercise 15.9. If \(y=a\cdot\tan^m(\theta^n)\), find the derivative of \(y\) with respect to \(\theta\).
Answer
\(amn\theta^{n-1} \tan^{m-1}\left(\theta^n\right)\sec^2 \theta^n\).
Solution
\[y=a \tan ^{m}\left(\theta^{n}\right)=a\left[\tan \left(\theta^{n}\right)\right]^{m}\] We write \[y=a u^{m}\quad\text{ where }\quad u=\tan v\ \text{ and }\ v=\theta^{n}.\] Then
\[\begin{align} & \frac{d y}{d \theta}=\frac{d y}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d \theta} \\ &=a m u^{m-1} \cdot \sec ^{2} v \cdot n \cdot \theta^{n-1} \\ &=a m (\tan v)^{m-1} \cdot \sec ^{2} v \cdot n \cdot \theta^{n-1} \\ &=a \cdot m \cdot n \cdot\left[\tan \left(\theta^{n}\right)\right]^{m-1} \cdot \sec ^{2}\left(\theta^{n}\right) \cdot \theta^{n-1} \end{align}\] Note that \(\sec ^{2}\left(\theta^{n}\right)\) means \(\left[\sec \left(\theta^{n}\right)\right]^{2}\) and \(\left[\tan \left(\theta^{n}\right)\right]^{m-1}\) can be written as \(\tan ^{m-1}\left(\theta^{n}\right)\). Hence \[\frac{d y}{d \theta}=a m n \theta^{n-1} \tan ^{m-1}\left(\theta^{n}\right) \sec ^{2}\left(\theta^{n}\right)\]
Exercise 15.10. If \(y=e^x \sin^2 x\), find \(\dfrac{dy}{dx}\) and \(\dfrac{d^2y}{dx^2}\).
Answer
\(e ^x \left(\sin^2 x + \sin2x\right)\);\(e ^x \left(\sin^2 x + 2\sin2x + 2\cos2x\right)\).
Solution
\[y=e^{x} \sin ^{2} x=e^{x}(\sin x)^{2}\]
\[\frac{d y}{d x}=\frac{d\left(e^{x}\right)}{d x} \sin ^{2} x+e^{x} \frac{d\left(\sin ^{2} x\right)}{d x}\]
To find \(\dfrac{d\left(\sin ^{2} x\right)}{d x}\), we notice that
\[\sin ^{2} x=(\sin x)^{2}\]
and
\[\begin{align} \frac{d\left(u^{2}\right)}{d x} & =\frac{d\left(u^{2}\right)}{d u} \cdot \frac{d u}{d x} \qquad(u=\sin x) \\ & =2 u \cdot \cos x \\ & =2 \sin x \cos x \\ & =\sin 2 x \end{align}\] Therefore \[\begin{align} \frac{d y}{d x} & =\frac{d\left(e^{x}\right)}{d x} \sin ^{2} x+e^{x} \frac{d\left(\sin ^{2} x\right)}{d x} \\ & =e^{x} \sin ^{2} x+e^{x} \sin 2 x \\ & =e^{x}\left(\sin ^{2} x+\sin 2 x\right) \end{align}\]
The second derivative: \[\begin{align} \frac{d y}{d x}&= \frac{d\left(e^{x}\left(\sin ^{2} x+\sin 2 x\right)\right)}{dx}\\ &=\frac{d(e^x)}{dx}\left(\sin ^{2} x+\sin 2 x\right)+e^x\left(\frac{d(\sin^2 x)}{dx}+\frac{d(\sin 2x)}{dx}\right)\\ &=e^x\left(\sin ^{2} x+\sin 2 x\right)+e^x\left(\underbrace{2\cos x \sin x}_{\sin 2x}+2\cos 2x\right)\\ &=e^x\left(\sin^2 x+2\sin 2x+2\cos 2x\right). \end{align}\]
Exercise 15.11. Differentiate the three equations of Exercises 14.II (see here), No. 4, and compare their derivatives, as to whether they are equal, or nearly equal, for very small values of \(x\), or for very large values of \(x\), or for values of \(x\) in the neighbourhood of \(x=b\).
Answer
\(\left(i\right) \dfrac{dy}{dx} = \dfrac{ab}{\left(x + b\right)^2}\); (ii) \(\dfrac{a}{b} e ^{-\frac{x}{b}}\); (iii) \(\dfrac{2ab}{\pi}\cdot\dfrac{1}{\left(b^2 + x^2\right)}\).
Solution
(i) Using the Quotient Rule: \[\frac{d y}{d x}=\frac{a(x+b)-a x}{(x+b)^{2}}=\frac{a b}{(x+b)^{2}}\]
(ii) \(\dfrac{d y}{d x}=-a \times\left(-\frac{1}{b}\right) e^{-\frac{x}{b}}=\frac{a}{b} e^{-\frac{x}{b}}\)
(iii) To differentiate \(y=\dfrac{2 a}{\pi} \arctan \left(\frac{x}{b}\right)\), we write \[y =\frac{2 a}{\pi} \arctan u \quad\text{where}\quad u=\frac{x}{b}.\] Then \[\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\frac{2 a}{\pi} \frac{1}{1+u^{2}} \cdot \frac{1}{b} \\ & =\frac{2 a}{\pi b} \frac{1}{1+\left(\frac{x}{b}\right)^{2}} \\ & =\frac{2 a b}{\pi} \frac{1}{x^{2}+b^{2}} \end{align}\]
When \(x\) is very large
\[\begin{align} & \frac{a b}{(x+b)^{2}} \approx \frac{a b}{x^{2}} \approx 0 \\ & \frac{a}{b} e^{-\frac{x}{b}} \approx \frac{a}{b} \times 0=0 \\ & \frac{2 a b}{x^{2}+b^{2}} \approx \frac{2 a b}{x^{2}} \approx 0 \end{align}\]
Therefore their slopes are almost zero for large values of \(x\).
When \(x \approx 0\) \[\begin{align} & \frac{a b}{(x+b)^{2}} \approx \frac{a b}{b^{2}}=\frac{a}{b} \\ & e^{-\frac{x}{b}} \approx 1 \Rightarrow \frac{a}{b} e^{-\frac{x}{b}} \approx \frac{a}{b} \\ & \frac{2 a b}{x^{2}+b^{2}} \approx \frac{2 a b}{b^{2}}=\frac{2 a}{b} \end{align}\] When \(x\) is small (\(x\approx 0\)), the slopes of \(y=\frac{a x}{x+b}\) and \(y=a\left(1-e^{-\frac{x}{b}}\right)\) are almost identical, but the slope of \(\frac{2 a}{\pi} \arctan \left(\frac{x}{b}\right)\) is twice as much as them.
When \(x \approx b\) \[\begin{align} & \frac{a b}{(x+b)^{2}} \approx \frac{a b}{(2 b)^{2}}=\frac{a}{4 b} \\ & \frac{a}{b} e^{-\frac{x}{b}} \approx \frac{a}{b} \cdot e^{-1}=\frac{a}{2.78 b} \\ & \frac{2 a b}{x^{2}+b^{2}} \approx \frac{2 a b}{2 b^{2}}=\frac{a}{b} . \end{align}\]
Exercise 15.12. Differentiate the following: \[\begin{align} \text{(i)}\quad y &= \sec x. & \text{(ii)}\quad y &= \arccos x. \\ \text{(iii)}\quad y &= \arctan x. & \text{(iv)}\quad y &= \text{arcsec} x. \\ \text{(v)}\quad y &= \tan x \times \sqrt{3 \sec x}. && \end{align}\]
Answer
(i) \(\dfrac{dy}{dx} = \sec x \tan x\);
(ii) \(\dfrac{dy}{dx} = - \dfrac{1}{\sqrt{ 1 - x^2}}\);
(iii) \(\dfrac{dy}{dx} = \dfrac{1}{ 1 + x^2}\);
(iv) \(\dfrac{dy}{dx} = \dfrac{1}{|x| \sqrt{ x^2 - 1}}\);
(v) \(\dfrac{dy}{dx} = \dfrac{\sqrt{ 3\sec x} \left(3\sec^2 x - 1\right)}{2}\).
Solution
(i) \(y=\sec x=\dfrac{1}{\cos x}\)
Using the Quotient Rule \[\begin{align} \frac{d y}{d x} & =\frac{-(-\sin x)}{\cos ^{2} x}=\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \\ & =\tan x \cdot \sec x . \end{align}\]
(ii) \(y=\arccos x\) (or \(y=\cos ^{-1} x\) )
If \(y=\arccos x\), then \(x=\cos y\) and \[\frac{d x}{d y}=-\sin y\]
\[\begin{align} \frac{d y}{d x} & =\frac{1}{\frac{d x}{d y}} \\ & =\frac{1}{-\sin y} \end{align}\] Since \(\sin y= \pm \sqrt{1-\cos ^{2} y}\) \[\frac{d y}{d x}=\frac{1}{-\sqrt{1-\cos ^{2} y}}\] Since \(x=\cos y\) \[\frac{d y}{d x}=\frac{1}{\mp \sqrt{1-x^{2}}}\] But which one is correct? The \(-\) sign or the \(+\) sign? If we look at the graph of \(y=\arccos x\), the slope is negative everywhere.
Hence \[\frac{d(\arccos x)}{d x}=\frac{-1}{\sqrt{1-x^{2}}}\]
(iii) If \(y=\arctan x\), then \(x=\tan y\) and
\[\frac{d x}{d y}=1+\tan ^{2} y \quad\left(\text { or } \sec ^{2} y\right)\] Therefore,
\[\begin{align} \frac{d y}{d x} & =\frac{1}{\dfrac{d x}{d y}} \\ & =\frac{1}{1+\tan ^{2} y} \\ & =\frac{1}{1+x^{2}} \end{align}\]
(iv) If \(y=\operatorname{arcsec} x\) then \(x=\sec y\). In part (i), we showed that
\[\frac{d x}{d y}=\tan y \cdot \sec y \text {. }\] Hence \[\begin{align} \frac{d y}{d x} & =\frac{1}{\dfrac{d x}{d y}} \\ & =\frac{1}{\tan y \cdot \sec y} \end{align}\] Since \(1+\tan ^{2} y=\sec ^{2} y \Rightarrow \tan y= \pm \sqrt{\sec ^{2} y-1}\), we have \[\begin{align} \frac{d y}{d x} & =\frac{1}{ \pm \sqrt{\sec ^{2} y-1} \cdot \sec y} \\ & =\frac{1}{ \pm x \sqrt{x^{2}-1}} . \end{align}\]
Now we need to decide about the sign.
As we can see from the graph \(y=\operatorname{arcsec} x\), the slope is always positive. So we must have
\[\begin{align} & \frac{d y}{d x}=\frac{1}{x \sqrt{x^{2}-1}} &&\text { if } x \geq 1 \\ & \frac{d y}{d x}=\frac{1}{-x \sqrt{x^{2}-1}} &&\text { if } x \leq-1 \end{align}\] We can combine these two and write \[\frac{d y}{d x}=\frac{1}{|x| \sqrt{x^{2}-1}}\]
(v) \(y=\tan x\times \sqrt{3 \sec x}=\sqrt{3}\, \tan x \cdot \sqrt{\sec x}\).
Using the Product Rule: \[\frac{d y}{d x}=\sqrt{3}\left[\frac{d(\tan x)}{d x} \cdot \sqrt{\sec x}+\tan x \frac{d(\sqrt{\sec x})}{d x}\right]\]
To find \(\dfrac{d\left(\sqrt{\sec x}\right)}{d x}\), let \(u=\sec x\). Then
\[\begin{align} \frac{d(\sqrt{u})}{d x} & =\frac{d(\sqrt{u})}{d u} \cdot \frac{d u}{d x} \\ & =\frac{1}{2 \sqrt{u}} \cdot \sec x \cdot \tan x \\ & =\frac{1}{2 \sqrt{\sec x}} \cdot \sec x \cdot \tan x \\ & =\frac{1}{2} \tan x\cdot \sqrt{\sec x} . \end{align}\]
Hence
\[\begin{align} \frac{d y}{d x} & =\sqrt{3}\left[\sec ^{2} x \sqrt{\sec x}+\frac{1}{2} \tan ^{2} x \sqrt{\sec x}\right] \\ & =\sqrt{3 \sec x} \quad\left[\sec ^{2} x+\frac{1}{2} \tan ^{2} x\right] \end{align}\]
We can simplify this further (or rewrite it in a different form) using \[1+\tan ^{2} x=\sec ^{2} x\]
\[\begin{align} & \frac{d y}{d x}=\sqrt{3 \sec x} \cdot\left[\sec ^{2} x+\frac{1}{2}\left(\sec ^{2} x-1\right)\right] \\ & =\frac{1}{2} \sqrt{3 \sec x}\left[3 \sec ^{2} x-1\right] \end{align}\]
Exercise 15.13. Differentiate \(y=\sin(2\theta +3)^{2.3}\).
Answer
\(\dfrac{dy}{d\theta} = 4.6\left(2\theta + 3\right)^{1.3} \cos\left(2\theta + 3\right)^{2.3}\).
Solution
\[\begin{align} & y=\sin (2 \theta+3)^{2.3}=\sin \left[(2 \theta+3)^{2.3}\right] \\ y & =\sin u, \quad u=v^{2.3}, \quad v=2 \theta+3 \\ \frac{d y}{d x}= & \frac{d y}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d x} \\ & =\cos u\left(2.3 v^{1.3}\right) \cdot 2 \\ & =4.6(2 \theta+3)^{1.3} \cos v^{2.3} \\ & =4.6(2 \theta+3)^{1.3} \cos \left[(2 \theta+3)^{2.3}\right] . \end{align}\]
Exercise 15.14. Differentiate \(y=\theta^3+3 \sin(\theta+3)-3^{\sin \theta} - 3^\theta\).
Answer
\(\dfrac{dy}{d\theta} = 3\theta^2 + 3\cos \left( \theta + 3 \right) - \ln 3 \left( \cos\theta \times 3^{\sin\theta} + 3\theta \right)\).
Solution
\[y=\theta^{3}+3 \sin (\theta+3)-3^{\sin \theta}-3^{\theta}\]
We differentiate term by term
\[\frac{d y}{d \theta}=3 \theta^{2}+3 \cos (\theta+3)-\frac{d\left(3^{\sin \theta}\right)}{d \theta}-3^{\theta} \cdot \ln 3\]
To find \(\dfrac{d\left(3^{\sin \theta}\right)}{d \theta}\), let \(\sin \theta=u\) and
\[\begin{align} \frac{d 3^{u}}{d \theta} & =\frac{d 3^{u}}{d u} \cdot \frac{d u}{d \theta} \\ & =3^{u} \cdot \ln 3 \cdot \cos \theta \\ & =3^{\sin \theta} \cdot \ln 3 \cdot \cos \theta \end{align}\]
Therefore
\[\frac{d y}{d \theta}=3 \theta^{2}+3 \cos (\theta+3)-\ln 3 \cos \theta 3^{\sin \theta}-\ln 3 \cdot 3^{\theta}\]
Exercise 15.15. Find the maximum or minimum of \(y=\theta \cos \theta\) (\(-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}\)).
Answer
\(\theta = \cot\theta; \theta = \pm 0.86\); \(y=\pm 0.56\) is max. for \(+\theta\), min. for \(-\theta\).
Solution
\[y=\theta \cos \theta\]
Using the Product Rule
\[\begin{align} \frac{d y}{d \theta}&=\cos \theta+\theta(-\sin \theta) \\ &=\cos \theta-\theta \sin \theta \\ \end{align}\]
\[\frac{d y}{d \theta}=\cos \theta-\theta \sin \theta=0 \Rightarrow \theta=\cot \theta\]
The solutions of \(\theta=\cot \theta\) can be obtained approximately
\[\theta \approx \pm 0.86\]
To distinguish between a maximum and a minimum, we use the Second Derivative Test:
\[\begin{align} \frac{d^{2} y}{d \theta^{2}} & =\frac{d(\cos \theta-\theta \sin \theta)}{d \theta} \\ & =-\sin \theta-(\sin \theta+\theta \cos \theta) \\ & =\theta \cos \theta-2 \sin \theta \end{align}\]
When \(\theta \approx 0.86\)
\[\frac{d^{2} y}{d \theta^{2}} \approx -0.95<0\] Hence, the curve is concave downward, and \(y\) has a maximum of \(0.86 \cos (0.86) \approx 0.56\) when \(\theta \approx 0.86\).
When \(\theta \approx-0.86\) \[\frac{d^{2} y}{d \theta^{2}} \approx 0.95>0\] Hence, the curve is concave upward, and thus \(y\) has a minimum of \(-0.86 \cos (-0.86) \approx-0.56\) when \(\theta \approx-0.86.\)