The Chain Rule
Sometimes one is stumped by finding that the expression to be differentiated is too complicated to tackle directly.
Thus, the equation \[y = (x^2+a^2)^{\frac{3}{2}}\] is awkward to a beginner.
Now the dodge to turn the difficulty is this: Write some symbol, such as \(u\), for the expression \(x^2 + a^2\); then the equation becomes \[y = u^{\frac{3}{2}},\] which you can easily manage; for \[\frac{dy}{du} = \frac{3}{2} u^{\frac{1}{2}}.\] Then tackle the expression \[u = x^2 + a^2,\] and differentiate it with respect to \(x\), \[\frac{du}{dx} = 2x.\] Then all that remains is plain sailing;
for \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx};\] that is, \[\begin{align} \frac{dy}{dx} &= \frac{3}{2} u^{\frac{1}{2}} \times 2x \\ &= \tfrac{3}{2} (x^2 + a^2)^{\frac{1}{2}} \times 2x \\ &= 3x(x^2 + a^2)^{\frac{1}{2}}; \end{align}\] and so the trick is done.
By and bye, when you have learned how to deal with sines, and cosines, and exponentials, you will find the chain rule of increasing usefulness.
Examples
Let us practice using the Chain Rule on a few examples.
Example 9.1. Differentiate \(y = \sqrt{a+x}\).
Solution. Let \(a+x = u\). \[ \frac{du}{dx} = 1;\quad y=u^{\frac{1}{2}};\quad \frac{dy}{du} = \tfrac{1}{2} u^{-\frac{1}{2}} = \tfrac{1}{2} (a+x)^{-\frac{1}{2}}.\] \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = \frac{1}{2\sqrt{a+x}}. \]
Example 9.2. Differentiate \(y = \dfrac{1}{\sqrt{a+x^2}}\).
Solution. Let \(a + x^2 = u\). \[ \frac{du}{dx} = 2x;\quad y=u^{-\frac{1}{2}};\quad \frac{dy}{du} = -\tfrac{1}{2}u^{-\frac{3}{2}}.\] \[\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx} = - \frac{x}{\sqrt{(a+x^2)^3}}. \]
Example 9.3. Differentiate \(y = \left(m - nx^{\frac{2}{3}} + \dfrac{p}{x^{\frac{4}{3}}}\right)^a\).
Solution. Let \(m - nx^{\frac{2}{3}} + px^{-\frac{4}{3}} = u\). \[ \frac{du}{dx} = -\tfrac{2}{3} nx^{-\frac{1}{3}} - \tfrac{4}{3} px^{-\frac{7}{3}};\] \[y = u^a;\quad \frac{dy}{du} = a u^{a-1}. \] \[\frac{dy}{dx} = \frac{dy}{du}\times\frac{du}{dx} = -a\left(m -nx^{\frac{2}{3}} + \frac{p}{x^{\frac{4}{3}}}\right)^{a-1} (\tfrac{2}{3} nx^{-\frac{1}{3}} + \tfrac{4}{3} px^{-\frac{7}{3}}). \]
Example 9.4. Differentiate \(y=\dfrac{1}{\sqrt{x^3 - a^2}}\).
Solution. Let \(u = x^3 - a^2\). \[\begin{align} \frac{du}{dx} &= 3x^2;\quad y = u^{-\frac{1}{2}};\quad \frac{dy}{du}=-\frac{1}{2}(x^3 - a^2)^{-\frac{3}{2}}. \\ \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx} = -\frac{3x^2}{2\sqrt{(x^3 - a^2)^3}}. \end{align}\]
Example 9.5. Differentiate \(y=\sqrt{\dfrac{1-x}{1+x}}\).
Solution. Write this as \(y=\dfrac{(1-x)^{\frac{1}{2}}}{(1+x)^{\frac{1}{2}}}\). \[\frac{dy}{dx} = \frac{(1+x)^{\frac{1}{2}}\, \dfrac{d(1-x)^{\frac{1}{2}}}{dx} - (1-x)^{\frac{1}{2}}\, \dfrac{d(1+x)^{\frac{1}{2}}}{dx}}{1+x}.\]
(We may also write \(y = (1-x)^{\frac{1}{2}} (1+x)^{-\frac{1}{2}}\) and differentiate as a product.)
Proceeding as in example 9.1 above, we get \[\frac{d(1-x)^{\frac{1}{2}}}{dx} = -\frac{1}{2\sqrt{1-x}}; \quad\text{and}\quad \frac{d(1+x)^{\frac{1}{2}}}{dx} = \frac{1}{2\sqrt{1+x}}.\]
Hence \[\begin{align} \frac{dy}{dx} &= - \frac{(1 + x)^{\frac{1}{2}}}{2(1 + x)\sqrt{1-x}} - \frac{(1 - x)^{\frac{1}{2}}}{2(1 + x)\sqrt{1+x}} \\ &= - \frac{1}{2\sqrt{1+x}\sqrt{1-x}} - \frac{\sqrt{1-x}}{2 \sqrt{(1+x)^3}};\\ \end{align}\] or \[\frac{dy}{dx} = - \frac{1}{(1+x)\sqrt{1-x^2}}.\]
Example 9.6. Differentiate \(y = \sqrt{\dfrac{x^3}{1+x^2}}\).
Solution. We may write this \[\begin{gathered} y = x^{\frac{3}{2}}(1+x^2)^{-\frac{1}{2}}; \\ \frac{dy}{dx} = \tfrac{3}{2} x^{\frac{1}{2}}(1 + x^2)^{-\frac{1}{2}} + x^{\frac{3}{2}} \times \frac{d\bigl[(1+x^2)^{-\frac{1}{2}}\bigr]}{dx}. \end{gathered}\]
Differentiating \((1+x^2)^{-\frac{1}{2}}\), as shown in example (2) above, we get \[\frac{d\bigl[(1+x^2)^{-\frac{1}{2}}\bigr]}{dx} = - \frac{x}{\sqrt{(1+x^2)^3}};\] so that \[\frac{dy}{dx} = \frac{3\sqrt{x}}{2\sqrt{1+x^2}} - \frac{\sqrt{x^5}}{\sqrt{(1+x^2)^3}} = \frac{\sqrt{x}(3+x^2)}{2\sqrt{(1+x^2)^3}}.\]
Example 9.7. Differentiate \(y=\left(x+\sqrt{x^2+x+a}\right)^3\).
Solution. Let \(x+\sqrt{x^2+x+a}=u\). \[\begin{gathered} \frac{du}{dx} = 1 + \frac{d\bigl[(x^2+x+a)^{\frac{1}{2}}\bigr]}{dx}. \\ y = u^3;\quad\text{and}\quad \frac{dy}{du} = 3u^2= 3\left(x+\sqrt{x^2+x+a}\right)^2. \end{gathered}\]
Now let \((x^2+x+a)^{\frac{1}{2}}=v\) and \((x^2+x+a) = w\). \[\begin{align} \frac{dw}{dx} &= 2x+1;\quad v = w^{\frac{1}{2}};\quad \frac{dv}{dw} = \tfrac{1}{2}w^{-\frac{1}{2}}. \\ \frac{dv}{dx} &= \frac{dv}{dw} \times \frac{dw}{dx} = \tfrac{1}{2}(x^2+x+a)^{-\frac{1}{2}}(2x+1). \end{align}\] Hence \[\begin{align} \frac{du}{dx} &= 1 + \frac{2x+1}{2\sqrt{x^2+x+a}}, \\ \frac{dy}{dx} &= \frac{dy}{du} \times \frac{du}{dx}\\ &= 3\left(x+\sqrt{x^2+x+a}\right)^2 \left(1 +\frac{2x+1}{2\sqrt{x^2+x+a}}\right). \end{align}\]
Example 9.8. Differentiate \(y=\sqrt{\dfrac{a^2+x^2}{a^2-x^2}} \sqrt[3]{\dfrac{a^2-x^2}{a^2+x^2}}\).
Solution. We get \[ y = \frac{(a^2+x^2)^{\frac{1}{2}} (a^2-x^2)^{\frac{1}{3}}} {(a^2-x^2)^{\frac{1}{2}} (a^2+x^2)^{\frac{1}{3}}} = (a^2+x^2)^{\frac{1}{6}} (a^2-x^2)^{-\frac{1}{6}}. \] \[\frac{dy}{dx} = (a^2+x^2)^{\frac{1}{6}} \frac{d\bigl[(a^2-x^2)^{-\frac{1}{6}}\bigr]}{dx} + \frac{d\bigl[(a^2+x^2)^{\frac{1}{6}}\bigr]}{(a^2-x^2)^{\frac{1}{6}}\, dx}. \]
Let \(u = (a^2-x^2)^{-\frac{1}{6}}\) and \(v = (a^2 - x^2)\). \[ u = v^{-\frac{1}{6}};\quad \frac{du}{dv} = -\frac{1}{6}v^{-\frac{7}{6}};\quad \frac{dv}{dx} = -2x.\] \[\frac{du}{dx} = \frac{du}{dv} \times \frac{dv}{dx} = \frac{1}{3}x(a^2-x^2)^{-\frac{7}{6}}. \]
Let \(w = (a^2 + x^2)^{\frac{1}{6}}\) and \(z = (a^2 + x^2)\). \[ w = z^{\frac{1}{6}};\quad \frac{dw}{dz} = \frac{1}{6}z^{-\frac{5}{6}};\quad \frac{dz}{dx} = 2x. \] \[\frac{dw}{dx} = \frac{dw}{dz} \times \frac{dz}{dx} = \frac{1}{3} x(a^2 + x^2)^{-\frac{5}{6}}. \]
Hence \[ \frac{dy}{dx} = (a^2+x^2)^{\frac{1}{6}} \frac{x}{3(a^2-x^2)^{\frac{7}{6}}} + \frac{x}{3(a^2-x^2)^{\frac{1}{6}} (a^2+x^2)^{\frac{5}{6}}}; \] or \[ \frac{dy}{dx} = \frac{x}{3} \left[\sqrt[6]{\frac{a^2+x^2}{(a^2-x^2)^7}} + \frac{1}{\sqrt[6]{(a^2-x^2)(a^2+x^2)^5]}} \right]. \]
Example 9.9. Differentiate \(y^n\) with respect to \(y^5\).
Solution.
\[\frac{d(y^n)}{d(y^5)} =\frac{\dfrac{d(y^n)}{dy}}{\dfrac{d(y^5)}{dy}}= \frac{ny^{n-1}}{5y^{5-1}} = \frac{n}{5} y^{n-5}.\]
Example 9.10. Find the first and second derivatives of \(y = \dfrac{x}{b} \sqrt{(a-x)x}\).
Solution. \[\frac{dy}{dx} = \frac{x}{b}\, \frac{d\bigl\{\bigl[(a-x)x\bigr]^{\frac{1}{2}}\bigr\}}{dx} + \frac{\sqrt{(a-x)x}}{b}.\tag{Product Rule}\]
Let \(\bigl[(a-x)x\bigr]^{\frac{1}{2}} = u\) and let \((a-x)x = w\); then \(u = w^{\frac{1}{2}}\). \[\frac{du}{dw} = \frac{1}{2} w^{-\frac{1}{2}} = \frac{1}{2w^{\frac{1}{2}}} = \frac{1}{2\sqrt{(a-x)x}}.\] \[\begin{align} &\frac{dw}{dx} = a-2x.\\ &\frac{du}{dw} \times \frac{dw}{dx} = \frac{du}{dx} = \frac{a-2x}{2\sqrt{(a-x)x}}. \end{align}\]
Hence \[\frac{dy}{dx} = \frac{x(a-2x)}{2b\sqrt{(a-x)x}} + \frac{\sqrt{(a-x)x}}{b} = \frac{x(3a-4x)}{2b\sqrt{(a-x)x}}.\]
Now \[\begin{align} \frac{d^2y}{dx^2} &= \frac{2b \sqrt{(a-x)x}\, (3a-8x) - \dfrac{(3ax-4x^2)b(a-2x)}{\sqrt{(a-x)x}}} {4b^2(a-x)x} \\ &= \frac{3a^2-12ax+8x^2}{4b(a-x)\sqrt{(a-x)x}}. \end{align}\]
(We shall need these two last derivatives later on. See exercise 11 from Chapter 12.)
Example 9.11. A cylinder whose height is twice the radius of the base is increasing in volume, so that all its parts keep always in the same proportion to each other; that is, at any instant, the cylinder is similar to the original cylinder. When the radius of the base is \(r\) feet, the surface area is increasing at the rate of \(20\) square inches per second; at what rate is its volume then increasing?1
Solution. \[\text{Area}= S = 2(\pi r^2)+ 2 \pi r \times 2r = 6 \pi r^2.\] \[\text{Volume} = V = \pi r^2 \times 2r=2 \pi r^3.\] \[\frac{dS}{dt} = 12\pi r\dfrac{dr}{dt}=20,\quad\Rightarrow\quad \frac{dr}{dt}=\frac{20}{12 \pi r},\] \[\frac{dV}{dt} = 6\pi r^2\, \frac{dr}{dt} = 6 \pi r^2 \times \frac{20}{12 \pi r} = 10r. \]
The volume changes at the rate of \(10r\) cubic inches.
Exercises I
Differentiate the following:
Exercise 9.1. \(y = \sqrt{x^2 + 1}\).
Answer
\(\dfrac{x}{\sqrt{ x^2 + 1}}\).
Solution
Let \(u=x^2+1\). Then \(y=u^{\frac{1}{2}}\) and \[\begin{align} \frac{d y}{d x}&=\frac{dy}{du}\cdot\frac{du}{dx}\\ &=\frac{1}{2}u^{\frac{1}{2}-1}\cdot (2 x)\\ &=x\left(x^{2}+1\right)^{-1 / 2}\\ &=\frac{x}{\sqrt{x^{2}+1}} \end{align}\]
Exercise 9.2. \(y = \sqrt{x^2+a^2}\).
Answer
\(\dfrac{x}{\sqrt{ x^2 + a^2}}\).
Solution
\[y=\sqrt{x^{2}+a^{2}}\]
Let \(u=x^2+a^2\). Then \(y=u^{\frac{1}{2}}\), and
\[\begin{align} \frac{d y}{d x}&=\frac{dy}{du}\cdot\frac{du}{dx}\\ &=\left(\frac{1}{2}u^{-\frac{1}{2}}\right)\cdot (2x)\\ &=\frac{1}{2}\left(x^{2}+a^{2}\right)^{\frac{1}{2}-1}(2 x)\\ &=\frac{x}{\sqrt{x^{2}+a^{2}}} \end{align}\]
Exercise 9.3. \(y = \dfrac{1}{\sqrt{a+x}}\).
Answer
\(- \dfrac{1}{2 \sqrt{(a + x)^3}}\).
Solution
\[\begin{align} & y=\frac{1}{\sqrt{a+x}}=(a+x)^{-1 / 2} \\ \Rightarrow &\frac{d y}{d x}=\left(-\frac{1}{ 2}\right)(a+x)^{-\frac{1}{2}-1}=\frac{-1}{2 \sqrt[2]{(a+x)^{3}}} \end{align}\]
Exercise 9.4. \(y = \dfrac{a}{\sqrt{a-x^2}}\).
Answer
\(\dfrac{ax}{\sqrt{(a - x^2)^3}}\).
Solution
\[y=\frac{a}{\sqrt{a-x^{2}}}=a\left(a-x^{2}\right)^{-\frac{1}{2}}\]
Let \(u=a-x^{2}\). Then
\[y=a u^{-\frac{1}{2}}\]
and
\[\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\left(-\frac{a}{2} u^{-\frac{3}{2}}\right)(-2 x) \\ & =\frac{a x}{u^{\frac{3}{2}}} \\ & =\frac{a x}{\sqrt{\left(a-x^{2}\right)^{3}}} \end{align}\]
Exercise 9.5. \(y = \dfrac{\sqrt{x^2-a^2}}{x^2}\).
Answer
\(\dfrac{2a^2 - x^2}{x^3 \sqrt{ x^2 - a^2}}\).
Solution
\[y=\frac{\sqrt{x^{2}-a^{2}}}{x^{2}}\] Using the Quotient Rule
\[\frac{d y}{d x}=\frac{\frac{d\left(\sqrt{x^{2}-a^{2}}\right)}{d x} \cdot x^{2}-2 x \sqrt{x^{2}-a^{2}}}{x^{4}}\]
To find \(\frac{d u}{d x}\) where \(u=\sqrt{x^2-a^2}\), let \(v=x^{2}-a^{2}\), then \(u=\sqrt{v}\) and
\[\begin{align} \frac{d u}{d x} & =\frac{d u}{d v} \cdot \frac{d v}{d x} \\ & =\frac{1}{2 \sqrt{v}} \cdot(2 x) \\ & =\frac{x}{\sqrt{x^{2}-a^{2}}} . \end{align}\] Therefore \[\frac{d\left(\sqrt{x^{2}-a^{2}}\right)}{d x}=\frac{x}{\sqrt{x^{2}-a^{2}}}\] and \[\begin{align} \frac{d y}{d x} & =\frac{\dfrac{x}{\sqrt{x^{2}-a^{2}}} \cdot x^{2}-2 x \sqrt{x^{2}-a^{2}}}{x^{4}} \\ & =\frac{x^{3}-2 x\left(x^{2}-a^{2}\right)}{x^{4} \sqrt{x^{2}-a^{2}}} \\ & =\frac{2 a x-x^{3}}{x^{4} \sqrt{x^{2}-a^{2}}} \\ & =\frac{2 a-x^{2}}{x^{3} \sqrt{x^{2}-a^{2}}} \end{align}\]
Exercise 9.6. \(y = \dfrac{\sqrt[3]{x^4+a}}{\sqrt{x^3+a}}\).
Answer
\(\dfrac{\frac{3}{2} x^2 \left[ \frac{8}{9} x \left( x^3 + a \right) - \left( x^4 + a \right) \right]}{(x^4 + a)^{\frac{2}{3}} (x^3 + a)^{\frac{3}{2}}}\)
Solution
To find \(\frac{d y}{d x}\), we need to find \(\frac{d\left(\sqrt[3]{x^{4}+a}\right)}{d x}\) and \(\frac{d\left(\sqrt{x^{3}+a}\right)}{d x}\).
\(u=\sqrt[3]{x^{4}+a}=\left(x^{4}+a\right)^{\frac{1}{3}}=v^{\frac{1}{3}}\) where \(v=x^{4}+a\). Then
\[\begin{align} \frac{d u}{d x} & =\frac{d u}{d u} \cdot \frac{d u}{d x} \\ & =\left(\frac{1}{3} v^{-\frac{2}{3}}\right)\left(4 x^{3}\right) \\ & =\frac{4 x^{3}}{3 \sqrt{\left(x^{2}+a\right)^{3}}} \end{align}\]
\(w=\sqrt{x^{3}+a}=\left(x^{3}+a\right)^{\frac{1}{2}}=z^{\frac{1}{2}}\), where \(z=x^{3}+a\). Then
\[\begin{align} \frac{d w}{d x} & =\frac{d w}{d z} \cdot \frac{d z}{d x} \\ & =\frac{1}{2} z^{-\frac{1}{2}} \cdot\left(3 x^{2}\right) \\ & =\frac{3 x^{2}}{2 z^{\frac{1}{2}}} \\ & =\frac{3 x^{2}}{2 \sqrt{x^{3}+a}} \end{align}\]
Now using the Quotient Rule:
\[\begin{align} \frac{d y}{d x} & =\frac{\frac{d\left(\sqrt[3]{x^{4}+a}\right)}{d x} \cdot \sqrt{x^{3}+a}-\frac{d\left(\sqrt{x^{3}+a}\right)}{d x} \sqrt[3]{x^{4}+a}}{x^{3}+a} \\ & =\frac{\frac{4 x^{3}}{3 \sqrt{\left(x^{4}+a\right)^{3}}} \sqrt{x^{3}+a}-\frac{3 x^{2}}{2 \sqrt{x^{3}+a}} \cdot \sqrt[3]{x^{4}+a}}{x^{3}+a} \\ & =\frac{\frac{4}{3} x^{3}\left(x^{3}+a\right)-\frac{3}{2} x^{2}\left(x^{4}+a\right)}{\sqrt{\left(x^{4}+a\right)^{3}}\left(x^{3}+a\right)^{\frac{3}{2}}} \\ & =\frac{\frac{3}{2} x^{2}\left[\frac{8}{9} x\left(x^{3}+a\right)-\left(x^{4}+a\right)\right]}{\left(x^{4}+a\right)^{\frac{3}{2}}\left(x^{3}+a\right)^{\frac{3}{2}}}. \end{align}\]
Exercise 9.7. \(y = \dfrac{a^2+x^2}{(a+x)^2}\).
Answer
\(\dfrac{2a \left(x - a \right)}{(x + a)^3}\).
Solution
Using the Quotient Rule: \[\begin{align} & \frac{d y}{d x}=\frac{2 x(a+x)^{2}-2(a+x)\left(a^{2}+x^{2}\right)}{(a+x)^{4}} \\ & \frac{d y}{d x}=\frac{2(a+x)\left[x(a+x)-\left(a^{2}+x^{2}\right)\right]}{(a+x)^{4}} \\ & \frac{d y}{d x}=\frac{2\left[a x+x^{2}-a^{2}-x^{2}\right]}{(a+x)^{3}}=\frac{2 a(x-a)}{(a+x)^{3}} \end{align}\]
Note that to find \(\dfrac{d\left((a+x)^2\right)}{dx}\), let \(u=a+x\). Then \[\frac{d\left((a+x)^2\right)}{dx}=\frac{d (u^2)}{du}\frac{du}{dx}=2u=2(a+x)\]
Exercise 9.8. Differentiate \(y^5\) with respect to \(y^2\).
Answer
\(\frac{5}{2} y^3\).
Solution
\[\frac{d \left(y^{5}\right)}{d \left(y^{2}\right)}=\frac{\dfrac{d\left(y^{5}\right)}{d y}}{\dfrac{d\left(y^{2}\right)}{d y}}=\frac{5 y^{4}}{2 y}=\frac{5}{2} y^{3}\]
Exercise 9.9. Differentiate \(y = \dfrac{\sqrt{1 - \theta^2}}{1 - \theta}\).
Answer
\(\dfrac{1}{(1 - \theta) \sqrt{1 - \theta^2}}\).
Solution
\[y=\frac{\sqrt{1-\theta^{2}}}{1-\theta}\]
To find \(\dfrac{dy}{d\theta}\), first we need to find the derivative of the numerator. To differentiate of \(\sqrt{1-\theta^2}\) with respect to \(\theta\), we can rewrite it as \(\left(1 - \theta^2\right)^{\frac{1}{2}}\) and apply the Chain Rule. Let \(u = 1-\theta^2\). Then: \[\begin{align} \frac{d\left( \left(1 - \theta^2\right)^{\frac{1}{2}}\right)}{dx}&=\frac{d \left(u^\frac{1}{2}\right)}{d\theta}\\ &=\frac{d \left(u^\frac{1}{2}\right)}{du}\frac{u}{d\theta}\\ &=\frac{1}{2} u^{-\frac{1}{2}} (-2\theta)\\ &=\frac{1}{2}(1-\theta^2)^{-\frac{1}{2}}(-2\theta)\\ &=-\frac{\theta}{\sqrt{1-\theta^2}} \end{align}\] Using the Quotient Rule
\[\begin{align} \frac{d y}{d \theta}&=\frac{-\dfrac{\theta}{\sqrt{1-\theta^2}(1-\theta)}-(-1) \sqrt{1-\theta^{2}}}{(1-\theta)^{2}}\\ &=\frac{-\theta(1-\theta)+(1-\theta^2)}{\sqrt{1-\theta^2}\ \ (1-\theta^2)}\\ &=\frac{1-\theta}{\sqrt{1-\theta^2}\ \ (1-\theta)^2}\\ &=\frac{1}{\sqrt{1-\theta^2}\ \ (1-\theta)}. \end{align}\]
Exercise 9.10. A spherical balloon is increasing in volume. If, when its radius is \(r\) feet, its volume is increasing at the rate of \(4\) cubic feet per second, at what rate is its surface then increasing?2
Answer
At the rate of \(\dfrac{8}{r}\) square feet per second.
Solution
The volume of the balloon is
\[V=\frac{4}{3} \pi r^{3}\]
and the surface of the balloon is
\[S=4 \pi r^{2}\]
We know
\[\frac{d V}{d t}=4~\frac{\mathrm{ft}^{3}}{\mathrm{~s}}\]
We want to find \(\dfrac{d S}{d t}\).
Differentiate both sides of the following equation with respect to time \(t\)
\[\frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}\]
Since \(\dfrac{d V}{d t}=4\), we have
\[\frac{d r}{d t}=\frac{1}{\pi r^{2}}\]
Now differentiate both sides of \(S=4 \pi r^{2}\), with respect to time \(t:\)
\[\frac{d S}{d t}=8 \pi r \frac{d r}{d t}\]
Substitute \(\frac{d r}{d t}=\frac{1}{\pi r^{2}}\) in the above equation gives
\[\frac{d S}{d t}=8 \pi r \frac{1}{\pi r^{2}}=\frac{8}{r} .\]
The process can be extended to three or more derivatives, so that \(\dfrac{dy}{dx} = \dfrac{dy}{dz} \times \dfrac{dz}{dv} \times \dfrac{dv}{dx}\).
Examples
Example 9.12. If \(z = 3x^4\);\(v = \dfrac{7}{z^2}\);\(y =\sqrt{1+v}\), find \(\dfrac{dv}{dx}\).
Solution. We have \[ \frac{dy}{dv} = \frac{1}{2\sqrt{1+v}};\quad \frac{dv}{dz} = -\frac{14}{z^3};\quad \frac{dz}{dx} = 12x^3. \] \[\frac{dy}{dx} = -\frac{168x^3}{(2\sqrt{1+v})z^3}= -\frac{28}{3x^5\sqrt{9x^8+7}}. \]
Example 9.13. If \(t = \dfrac{1}{5\sqrt{\theta}}\);\(x = t^3 + \dfrac{t}{2}\);\(v = \dfrac{7x^2}{\sqrt[3]{x-1}}\), find \(\dfrac{dv}{d\theta}\).
Solution. Since \[\frac{dv}{d\theta}=\frac{dv}{dx}\cdot\frac{dx}{dt}\cdot\frac{dt}{d\theta}\] to calculate \(\dfrac{dv}{d\theta}\), we first need to find \(\dfrac{dv}{dx}\), \(\dfrac{dx}{dt}\), and \(\dfrac{dt}{d\theta}\).
Differentiating \(v\) with respect to \(x\) gives \[\frac{dv}{dx}=\frac{\sqrt[3]{x-1}\dfrac{d(7x^{2})}{dx}-7x^{2}\dfrac{d(\sqrt[3]{x-1})}{dx}}{\left(\sqrt[3]{x-1}\right)^{2}} \tag{Quotient Rule}\] Placing \(\dfrac{d(7x^2)}{dx}=14x\) and \(\dfrac{d\left(\sqrt[3]{x-1})\right)}{dx}=\dfrac{d\left((x-1)^{\frac{1}{3}}\right)}{dx}=\frac{1}{3}(x-1)^{-\frac{2}{3}}\) into the above expression, we get \[\frac{dv}{dx}=\frac{\sqrt[3]{x-1}(14x)-\frac{7}{3}x^{2}(x-1)^{-\frac{2}{3}}}{\left(\sqrt[3]{x-1}\right)^{2}}\] To simplify, multiply both the numerator and denominator by \(3(x-1)^{\frac{2}{3}}\): \[\begin{align} \frac{dv}{dx} &=\frac{7x\left[2(x-1)^{\frac{1}{3}}-\frac{1}{3}x(x-1)^{-\frac{2}{3}}\right]}{(x-1)^{\frac{2}{3}}}\times\frac{3(x-1)^{\frac{2}{3}}}{3(x-1)^{\frac{2}{3}}}\\[9pt] &=\frac{7x\left[6(x-1)-x\right]}{3(x-1)^{\frac{4}{3}}}\\ &=\frac{7x(5x-6)}{3\sqrt[3]{(x-1)^4}}. \end{align}\]
\[x = t^3 + \dfrac{t}{2}\quad\Rightarrow \quad \frac{dx}{dt}=3t^2+\frac{1}{2}\]
\[t = \dfrac{1}{5\sqrt{\theta}}=\frac{1}{5}\theta^{-\frac{1}{2}}\quad\Rightarrow\quad\dfrac{dt}{d\theta}=\frac{1}{5}\times\left(-\frac{1}{2}\right)\theta^{-\frac{3}{2}}=-\frac{1}{10\sqrt{\theta^3}}\]
So \[\begin{gathered} \frac{dv}{dx} = \frac{7x(5x-6)}{3\sqrt[3]{(x-1)^4}};\quad \frac{dx}{dt} = 3t^2 + \tfrac{1}{2};\quad \frac{dt}{d\theta} = -\frac{1}{10\sqrt{\theta^3}}. \end{gathered}\] Hence \[\begin{gathered} \frac{dv}{d\theta} = -\frac{7x(5x-6)(3t^2+\frac{1}{2})} {30\sqrt[3]{(x-1)^4} \sqrt{\theta^3}}, \end{gathered}\] an expression in which \(x\) must be replaced by its value, and \(t\) by its value in terms of \(\theta\).
Example 9.14. If \(\theta = \dfrac{3a^2x}{\sqrt{x^3}}\);\(\omega = \dfrac{\sqrt{1-\theta^2}}{1+\theta}\);and \(\phi = \sqrt{3} - \dfrac{1}{\omega\sqrt{2}}\), find \(\dfrac{d\phi}{dx}\).
Solution. We get \[\theta = 3a^2x^{-\frac{1}{2}};\quad \omega = \frac{\sqrt{(1-\theta)(1+\theta)}}{1+\theta}=\sqrt{\frac{1-\theta}{1+\theta}};\quad \text{and}\quad \phi = \sqrt{3} {-} \frac{1}{\sqrt{2}} \omega^{-1}.\] \[\frac{d\theta}{dx} = -\frac{3a^2}{2\sqrt{x^3}};\quad \frac{d\omega}{d\theta} = -\frac{1}{(1+\theta)\sqrt{1-\theta^2}}\] (see example 9.5); and \[\frac{d\phi}{d\omega} = \frac{1}{\sqrt{2}\omega^2}.\]
So that \(\dfrac{d\theta}{dx} = \dfrac{1}{\sqrt{2} \times \omega^2} \times \dfrac{1}{(1+\theta) \sqrt{1-\theta^2}} \times \dfrac{3a^2}{2\sqrt{x^3}}\).
Replace now first \(\omega\), then \(\theta\) by its value.
Exercises II
Exercise 9.11. If \(u = \frac{1}{2}x^3\);\(v = 3(u+u^2)\); and \(w = \dfrac{1}{v^2}\), find \(\dfrac{dw}{dx}\).
Answer
\(\dfrac{dw}{dx} = -\dfrac{x^2 \left( 1 + x^3 \right)} {3 \left(\frac{1}{2} x^3 + \frac{1}{4} x^6 \right)^3}\).
Solution
\[\begin{align} & \frac{d w}{d x}=\frac{d w}{d v} \cdot \frac{d v}{d u} \cdot \frac{d u}{d x} \\ & =\frac{d\left(v^{-2}\right)}{d v} \cdot \frac{d\left(3 u+3 u^{2}\right)}{d u} \cdot \frac{d\left(\frac{1}{2} x^{3}\right)}{d x} \\ & =\left(-2 v^{-3}\right)(3+6 u)\left(\frac{3}{2} x^{2}\right) \\ & =\left(-\frac{2}{v^3}\right)(3+6 u)\left(\frac{3}{2} x^{2}\right) \\ & =\frac{-2}{\left(3 u+3 u^{2}\right)^{3}}(3+6 u)\left(\frac{3}{2} x^{2}\right) \\ & =\frac{-6(1+2 u)}{27\left(u+u^{2}\right)^{3}} \times \frac{3}{2} x^{2} \\ & =-\frac{1}{3} \frac{\left(1+2 \times \frac{1}{2} x^{3}\right)}{\left(\frac{1}{2} x^{3}+\frac{1}{4} x^{6}\right)^{3}} x^{2} \\ & =-\frac{x^{2}}{3} \frac{\left(1+x^{3}\right)}{\left(\frac{1}{2} x^{3}+\frac{1}{4} x^{6}\right)^{3}} \end{align}\]
Exercise 9.12. If \(y = 3x^2 + \sqrt{2}\);\(z = \sqrt{1+y}\); and \(v = \dfrac{1}{\sqrt{3}+4z}\), find \(\dfrac{dv}{dx}\).
Answer
\(\dfrac{dv}{dx} = - \dfrac{12x}{\sqrt{1 + \sqrt{2} + 3x^2} \left(\sqrt{3} + 4 \sqrt{1 + \sqrt{2} + 3x^2}\right)^2}\).
Solution
\[y=3 x^{2}+\sqrt{2}, \quad z=\sqrt{1+y},\quad v=\frac{1}{\sqrt{3}+4 z}.\]
\[\frac{d y}{d x}=6 x,\] \[\begin{align} \frac{d z}{d y}&=\frac{1}{2}(1+y)^{\frac{1}{2}-1}\\ &=\frac{1}{2 \sqrt{1+y}}\\ &=\frac{1}{2 \sqrt{1+3 x^{2}+\sqrt{2}}}\\ &=\frac{1}{2 \sqrt{1+\sqrt{2}+3 x^{2}}} \end{align}\] \[\begin{align} \frac{d v}{d z}&=\frac{-4}{(\sqrt{3}+4 z)^{2}}\\ &=\frac{-4}{(\sqrt{3}+4 \sqrt{1+y})^{2}}\\ &=\frac{-4}{\left(\sqrt{3}+4 \sqrt{1+\sqrt{2}+3 x^{2}}\right)^{2}} \end{align}\]
\[\begin{align} \frac{d v}{d x}&=\frac{d v}{d z} \cdot \frac{d z}{d y} \cdot \frac{d y}{d x}= \\ & =\frac{-4}{\left(\sqrt{3}+4 \sqrt{1+\sqrt{2}+3 x^{2}}\right)^{2}} \cdot \frac{1}{2 \sqrt{1+\sqrt{2}+3 x^{2}}} \cdot 6 x \\ & =\frac{-12 x}{\sqrt{1+\sqrt{2}+3 x^{2}}\left(\sqrt{3}+4 \sqrt{1+\sqrt{2}+3 x^{2}}\right)^{2}} \\ \end{align}\]
Exercise 9.13. If \(y = \dfrac{x^3}{\sqrt{3}}\);\(z = (1+y)^2\); and \(u = \dfrac{1}{\sqrt{1+z}}\), find \(\dfrac{du}{dx}\).
Answer
\(\dfrac{du}{dx} = - \dfrac{x^2 \left(\sqrt{3} + x^3 \right)} {\sqrt{ \left[ 1 + \left( 1 + \dfrac{x^3}{\sqrt{3}} \right) ^2 \right]^3}}\).
Solution
\[y =\frac{x^{3}}{\sqrt{3}} \Rightarrow \frac{d y}{d x}=\frac{3}{\sqrt{3}} x^{2}=\sqrt{3} x^{2}\]
\[z =(1+y)^{2} \Rightarrow \frac{d z}{d y}=2(1+y)\] \[u =\frac{1}{\sqrt{1+z}}=(1+z)^{-\frac{1}{2}} \Rightarrow \frac{d u}{d z}=-\frac{1}{2}(1+z)^{-\frac{3}{2}}\]
\[\begin{align} \frac{d u}{d x} & =\frac{d u}{d z} \cdot \frac{d z}{d y} \cdot \frac{d y}{d x} \\ & =-\frac{1}{2(1+z)^{\frac{3}{2}}} \cdot 2(1+y) \times \sqrt{3} x^{2} \\ & =-\frac{\sqrt{3} x^{2}}{\left(1+(1+y)^{2}\right)^{\frac{3}{2}}}(1+y) \\ & =-\frac{\sqrt{3} x^{2}}{\left(1+\left(1+\dfrac{x^{3}}{\sqrt{3}}\right)^{2}\right)^{\frac{3}{2}}}\left(\sqrt{3}+x^{3}\right) \\ & =-\frac{\sqrt{3} x^2}{\left(1+\left(1+\dfrac{x^3}{\sqrt{3}}\right)^2\right)^{\frac{3}{2}}}\left(1+\frac{x^3}{\sqrt{3}}\right) \\ & =-\frac{x^2\left(\sqrt{3}+x^3\right)}{\sqrt{\left[1+\left(1+\dfrac{x^3}{\sqrt{3}}\right)^2\right]^3}} \end{align}\]