Derivatives Of Integrals

Differentiating an integral whose limits depend on x requires nothing new: the Fundamental Theorem handles a variable upper limit, and the Chain Rule handles everything else.

Formula	Name	Condition
$\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$	Basic Rule	$a$ constant, $f$ continuous
\dfrac{d}{dx}\displaystyle\int_{a}^{g(x)}f(t)\,dt=f\!\left(g(x)\right)g'(x)	Variable Upper Limit	$g$ differentiable
\dfrac{d}{dx}\displaystyle\int_{h(x)}^{b}f(t)\,dt=-f\!\left(h(x)\right)h'(x)	Variable Lower Limit	Sign from the reversal rule
\dfrac{d}{dx}\displaystyle\int_{h(x)}^{g(x)}f(t)\,dt=f\!\left(g(x)\right)g'(x)-f\!\left(h(x)\right)h'(x)	Leibniz Rule	Both limits variable

In the previous section, we learned that the Fundamental Theorem of Calculus says that if $f$ is continuous on $[a, x]$ , then

\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)

The colouring is worth pausing on. The variable of integration t is a dummy and never appears in the answer; the variable that survives is the one in the upper limit, and the answer is the integrand evaluated at it.

The Basic Rule

Verify that

\frac{d}{d x} \int_{1}^{x} f (t) d t = f (x)

for $f (t) = t^{2}$ .

Solution

First, let us find $\int_{1}^{x} f (t) d t$ :

\begin{aligned} \int_{1}^{x}t^{2}\,dt &=\left.\frac{t^{3}}{3}\right]_{1}^{x}\\ &=\frac{x^{3}}{3}-\frac{1}{3}. \end{aligned}

Now we differentiate the result:

\frac{d}{d x} (\frac{x^{3}}{3} - \frac{1}{3}) = x^{2} = f (x) .

So the formula holds. Notice how the constant lower limit contributed only the constant $- \frac{1}{3}$ , which vanished on differentiation. That is exactly why the value of $a$ never appears in the answer.

Let $F (x) = \int_{- 1}^{x} \frac{1}{2 + 2 t + t^{2}} d t$ . Find F'(3).

Solution

Because $t^{2} + 2 t + 2 = (t + 1)^{2} + 1$ is never zero, the integrand

f (t) = \frac{1}{2 + 2 t + t^{2}}

is continuous everywhere, in particular on $[- 1, 3]$ .

[Recall that a rational function $R (x) = \frac{P (x)}{Q (x)}$ , where $P$ and $Q$ are polynomials, is continuous at $x = a$ provided $Q (a) \neq 0$ .]

Therefore, we can apply the Fundamental Theorem of Calculus and conclude

F'(3)=f(3)=\frac{1}{2+2(3)+3^{2}}=\frac{1}{17}.

Note that we never had to find $F$ itself, which would have required an antiderivative of the integrand. This is the practical power of Part 1: it gives the derivative without the antiderivative.

Combining with the Chain Rule

We may use the Chain Rule in conjunction with the first part of the Fundamental Theorem of Calculus to find the derivative when the upper limit or the lower limit of integration is a function rather than the bare variable x. For example, if

F (x) = \int_{a}^{g (x)} f (t) d t,

then to find F'(x) we let $u = g (x)$ . Then

\begin{aligned} \frac{dF}{dx} &=\frac{dF}{du}\frac{du}{dx}\\ &=\left(\frac{d}{du}\int_{a}^{u}f(t)\,dt\right)g'(x)\\ &=f(u)\,g'(x) &&{\small(\text{by the Fundamental Theorem of Calculus})}\\ &=f\!\left(g(x)\right)g'(x). \end{aligned}

Rule of thumb. Substitute the upper limit into the integrand, then multiply by the derivative of that upper limit.

Find S'(x), if

S (x) = \int_{0}^{x^{2}} \frac{\sin t}{t} d t .

Solution

Let $u = x^{2}$ . Then

\begin{aligned} S'(x) &=\left(\frac{d}{du}\int_{0}^{u}\frac{\sin t}{t}\,dt\right)\frac{du}{dx}\\ &=\left(\frac{\sin u}{u}\right)(2x)\\ &=\frac{\sin x^{2}}{x^{2}}(2x) &&{\small(u=x^{2})}\\ &=\frac{2\sin x^{2}}{x}. \end{aligned}

The integrand $\frac{\sin t}{t}$ has no elementary antiderivative, so there is no way to evaluate $S (x)$ in closed form. The derivative, however, costs one line.

When Both Limits Are Variable

If both limits depend on x, split the integral at an arbitrary constant c and treat the two pieces separately, using the reversal rule on the piece whose variable limit is at the bottom.

Find

\frac{d}{d x} (\int_{x^{2}}^{x^{3}} \ln t d t), x > 0.

Solution

We write

\begin{aligned} \int_{x^{2}}^{x^{3}}\ln t\,dt &=\int_{x^{2}}^{c}\ln t\,dt+\int_{c}^{x^{3}}\ln t\,dt\\ &=-\int_{c}^{x^{2}}\ln t\,dt+\int_{c}^{x^{3}}\ln t\,dt, \end{aligned}

where $c$ is an arbitrary constant. Because the derivative of a sum is the sum of the derivatives,

\frac{d}{d x} (\int_{x^{2}}^{x^{3}} \ln t d t) = - \frac{d}{d x} (\int_{c}^{x^{2}} \ln t d t) + \frac{d}{d x} (\int_{c}^{x^{3}} \ln t d t) .

Now we differentiate each piece as in the previous example:

\begin{aligned} \frac{d}{dx}\left(\int_{c}^{x^{2}}\ln t\,dt\right) &=\left(\frac{d}{du}\int_{c}^{u}\ln t\,dt\right)\frac{du}{dx} &&{\small(u=x^{2})}\\ &=(\ln u)(2x)\\ &=\left(\ln x^{2}\right)(2x)\\ &=2x\ln x^{2}, \end{aligned}\begin{aligned} \frac{d}{dx}\left(\int_{c}^{x^{3}}\ln t\,dt\right) &=\left(\frac{d}{dv}\int_{c}^{v}\ln t\,dt\right)\frac{dv}{dx} &&{\small(v=x^{3})}\\ &=(\ln v)\left(3x^{2}\right)\\ &=3x^{2}\ln x^{3}. \end{aligned}

Therefore,

\frac{d}{d x} (\int_{x^{2}}^{x^{3}} \ln t d t) = - 2 x \ln x^{2} + 3 x^{2} \ln x^{3} .

We may simplify further using $\ln x^{r} = r \ln x$ :

\begin{aligned} \frac{d}{dx}\left(\int_{x^{2}}^{x^{3}}\ln t\,dt\right) &=-4x\ln x+9x^{2}\ln x\\ &=x(9x-4)\ln x. \end{aligned}

Notice that the arbitrary constant $c$ disappeared, as it had to: the original integral does not depend on it.

An Application to Indeterminate Limits

Find

lim_{x \to 0^{+}} \frac{\int_{0}^{4 x^{2}} \sin \sqrt{t} d t}{x^{3}} .

Solution

Because as $x \to 0$ ,

\int_{0}^{4 x^{2}} \sin \sqrt{t} d t \to 0, x^{3} \to 0,

we have an indeterminate limit of type $0 / 0$ . So we can apply l'Hôpital's rule:

lim_{x \to 0^{+}} \frac{\int_{0}^{4 x^{2}} \sin \sqrt{t} d t}{x^{3}} \overset{H}{=} lim_{x \to 0^{+}} \frac{\frac{d}{d x} \int_{0}^{4 x^{2}} \sin \sqrt{t} d t}{\frac{d}{d x} x^{3}} .

The numerator's derivative comes from the Chain Rule combined with the Fundamental Theorem:

\begin{aligned} \frac{d}{dx}\int_{0}^{4x^{2}}\sin\sqrt{t}\,dt &=\left(\frac{d}{du}\int_{0}^{u}\sin\sqrt{t}\,dt\right)\frac{du}{dx} &&{\small(u=4x^{2})}\\ &=\left(\sin\sqrt{u}\right)(8x)\\ &=8x\sin\sqrt{4x^{2}}. \end{aligned}

Therefore

\begin{aligned} \lim_{x\to0^{+}}\frac{\frac{d}{dx}\int_{0}^{4x^{2}}\sin\sqrt{t}\,dt}{\frac{d}{dx}x^{3}} &=\lim_{x\to0^{+}}\frac{8x\sin\sqrt{4x^{2}}}{3x^{2}}\\ &=\lim_{x\to0^{+}}\frac{8x\sin2x}{3x^{2}} &&{\small(\sqrt{4x^{2}}=|2x|=2x\text{ since }x>0)}\\ &=\lim_{x\to0^{+}}\frac{8}{3}\cdot\frac{\sin2x}{x}\\ &\stackrel{H}{=}\frac{8}{3}\lim_{x\to0^{+}}\frac{2\cos2x}{1}\\ &=\frac{16}{3}. \end{aligned}

Implicit Differentiation of Integrals

Find $d y / d x$ if $y$ is implicitly defined by the equation

\int_{0}^{y} e^{- t^{2}} d t + \int_{0}^{x^{2}} \sin t d t = 0.

Solution

Using implicit differentiation, we get

\frac{d}{d x} (\int_{0}^{y} e^{- t^{2}} d t) + \frac{d}{d x} (\int_{0}^{x^{2}} \sin t d t) = 0,

\frac{d}{d y} (\int_{0}^{y} e^{- t^{2}} d t) \frac{d y}{d x} + \frac{d}{d x} (\int_{0}^{x^{2}} \sin t d t) = 0,

\begin{aligned} e^{-y^{2}}\frac{dy}{dx}+\frac{d}{dx}\left(\int_{0}^{x^{2}}\sin t\,dt\right)=0 &&{\small(\text{Fundamental Theorem of Calculus})} \end{aligned}\begin{aligned} e^{-y^{2}}\frac{dy}{dx}+\frac{d}{du}\left(\int_{0}^{u}\sin t\,dt\right)\frac{du}{dx}=0 &&{\small(u=x^{2})} \end{aligned}

e^{- y^{2}} \frac{d y}{d x} + \sin u \frac{d u}{d x} = 0

\begin{aligned} e^{-y^{2}}\frac{dy}{dx}+\left(\sin x^{2}\right)(2x)=0 &&{\small(u=x^{2})} \end{aligned}\begin{aligned} \frac{dy}{dx} &=-\frac{1}{e^{-y^{2}}}\,2x\sin x^{2}\\ &=-2x\,e^{y^{2}}\sin x^{2}. \end{aligned}

Neither integral in the original equation has an elementary antiderivative, so there is no hope of solving for $y$ explicitly. Implicit differentiation combined with Part 1 sidesteps that difficulty completely.

Exercises

Find F'(x) for each:

(a) $F (x) = \int_{2}^{x} \sqrt{1 + t^{4}} d t$

(b) $F (x) = \int_{0}^{x} \frac{d t}{1 + t^{2} + t^{4}}$

(c) $F (x) = \int_{x}^{5} \cos (t^{2}) d t$

Answer

(a) $\sqrt{1 + x^{4}}$ (b) $\frac{1}{1 + x^{2} + x^{4}}$ (c) $- \cos (x^{2})$

Solution

(a) and (b) are direct applications of Part 1: substitute $x$ for $t$ in the integrand.

(c) The variable is in the lower limit. Reverse the limits first:

F (x) = \int_{x}^{5} \cos (t^{2}) d t = - \int_{5}^{x} \cos (t^{2}) d t,

so F'(x)=-\cos(x^{2}). The sign is the whole content of this part; a variable lower limit means the accumulated area shrinks as $x$ increases.

Find $\frac{d y}{d x}$ if

(a) $y = \int_{1}^{x^{3}} \frac{d t}{t}$ ( $x > 0$ )

(b) $y = \int_{0}^{\sin x} \sqrt{1 + t^{2}} d t$

Answer

(a) $\frac{3}{x}$ (b) $\cos x \sqrt{1 + \sin^{2} x}$

Solution

(a) With $u = x^{3}$ , $\frac{d y}{d x} = \frac{1}{u} \cdot 3 x^{2} = \frac{3 x^{2}}{x^{3}} = \frac{3}{x}$ .

Check. Evaluating first, $y = \ln x^{3} = 3 \ln x$ , whose derivative is indeed $3 / x$ .

(b) With $u = \sin x$ , $\frac{d y}{d x} = \sqrt{1 + u^{2}} \cdot \cos x = \cos x \sqrt{1 + \sin^{2} x}$ .

Find $\frac{d}{d x} \int_{x}^{x^{2}} \frac{d t}{\sqrt{1 + t^{4}}}$ .

Answer

$\frac{2 x}{\sqrt{1 + x^{8}}} - \frac{1}{\sqrt{1 + x^{4}}}$

Solution

Both limits vary, so use the Leibniz Rule with $g (x) = x^{2}$ and $h (x) = x$ :

\frac{d}{dx}\int_{h(x)}^{g(x)}f(t)\,dt=f\!\left(g(x)\right)g'(x)-f\!\left(h(x)\right)h'(x).

Here $f (t) = \frac{1}{\sqrt{1 + t^{4}}}$ , so $f (x^{2}) = \frac{1}{\sqrt{1 + x^{8}}}$ and g'(x)=2x, while $f (x) = \frac{1}{\sqrt{1 + x^{4}}}$ and h'(x)=1. Substituting gives the answer.

Let $F (x) = \int_{0}^{x} (t^{2} - 4) d t$ . Find the intervals where $F$ is increasing and decreasing, and locate its local extremum.

Answer

Increasing on $(- \infty, - 2)$ and $(2, \infty)$ , decreasing on $(- 2, 2)$ ; local maximum at $x = - 2$ , local minimum at $x = 2$ .

Solution

By Part 1, F'(x)=x^{2}-4=(x-2)(x+2), which is positive when $| x | > 2$ and negative when $| x | < 2$ .

So $F$ increases on $(- \infty, - 2)$ , decreases on $(- 2, 2)$ , and increases again on $(2, \infty)$ . By the First Derivative Test, $x = - 2$ gives a local maximum and $x = 2$ a local minimum.

The lesson is that the graph of the integrand tells us everything about the shape of the accumulation function: $F$ grows where $f$ is positive and shrinks where $f$ is negative.

Let $G (x) = \int_{1}^{x} \frac{d t}{t}$ for $x > 0$ . Show that $G$ is increasing and concave down on $(0, \infty)$ , and identify $G$ .

Answer

G'(x)=\dfrac{1}{x}>0 and G''(x)=-\dfrac{1}{x^{2}}<0; $G (x) = \ln x$ .

Solution

Part 1 gives G'(x)=\dfrac{1}{x}, which is positive for $x > 0$ , so $G$ is increasing. Differentiating again, G''(x)=-\dfrac{1}{x^{2}}<0, so the graph is concave down throughout.

Since G'(x)=1/x and $G (1) = \int_{1}^{1} = 0$ , the function $G$ is the antiderivative of $1 / x$ that vanishes at 1, which is exactly $\ln x$ . Many treatments take this integral as the definition of the natural logarithm and derive its properties from Part 1.

Evaluate $lim_{x \to 0} \frac{1}{x} \int_{0}^{x} \frac{d t}{1 + t^{3}}$ .

Answer

$1$

Solution

Both numerator and denominator tend to 0, so the limit is of type $0 / 0$ and l'Hôpital's rule applies:

lim_{x \to 0} \frac{\int_{0}^{x} \frac{d t}{1 + t^{3}}}{x} \overset{H}{=} lim_{x \to 0} \frac{\frac{1}{1 + x^{3}}}{1} = 1.

Interpreted differently, the expression is the average value of $\frac{1}{1 + t^{3}}$ over $[0, x]$ , and as the interval shrinks to a point that average approaches the value of the function at 0, which is 1.

Find $lim_{x \to 0} \frac{\int_{0}^{x^{2}} \sqrt{1 + t^{3}} d t}{x^{2}}$ .

Answer

$1$

Solution

Again of type $0 / 0$ . Differentiating the numerator requires the Chain Rule with $u = x^{2}$ :

\frac{d}{d x} \int_{0}^{x^{2}} \sqrt{1 + t^{3}} d t = \sqrt{1 + x^{6}} \cdot 2 x .

The denominator differentiates to $2 x$ . Hence

lim_{x \to 0} \frac{\sqrt{1 + x^{6}} \cdot 2 x}{2 x} = lim_{x \to 0} \sqrt{1 + x^{6}} = 1.

Find $\frac{d y}{d x}$ if $y$ is defined implicitly by

\int_{0}^{y} \sqrt{1 + t^{2}} d t + \int_{0}^{x} \cos t d t = 1.

Answer

$\frac{d y}{d x} = - \frac{\cos x}{\sqrt{1 + y^{2}}}$

Solution

Differentiate both sides with respect to $x$ . For the first integral, the variable limit is $y$ , so the Chain Rule contributes $d y / d x$ :

\sqrt{1 + y^{2}} \frac{d y}{d x} + \cos x = 0.

Solving,

\frac{d y}{d x} = - \frac{\cos x}{\sqrt{1 + y^{2}}} .

The right-hand constant 1 differentiates to zero and plays no role beyond fixing which curve we are on.

Find the equation of the tangent line to the curve $y = \int_{2}^{x} \frac{t d t}{\sqrt{t^{2} + 5}}$ at the point where $x = 2$ .

Answer

$y = \frac{2}{3} (x - 2)$

Solution

At $x = 2$ the limits coincide, so $y (2) = \int_{2}^{2} = 0$ ; the point of tangency is $(2, 0)$ .

By Part 1,

y'(x)=\frac{x}{\sqrt{x^{2}+5}},\qquad y'(2)=\frac{2}{\sqrt{9}}=\frac{2}{3}.

The point-slope form gives $y - 0 = \frac{2}{3} (x - 2)$ .

Note how little work this took: no antiderivative was ever computed, though in this case one exists.

Explain the error in the following calculation and give the correct answer:

\frac{d}{d x} \int_{0}^{x} x t d t = x \cdot x = x^{2} .

Answer

The factor $x$ inside the integral is constant with respect to $t$ and must be pulled out first; the correct derivative is $\frac{3 x^{2}}{2}$ .

Solution

Part 1 applies to $\int_{0}^{x} f (t) d t$ , where the integrand depends on the dummy variable $t$ only. Here the integrand $x t$ also contains the endpoint variable $x$ , so the theorem cannot be applied in that form.

The fix is to treat $x$ as a constant during the integration, since the integration is over $t$ :

\int_{0}^{x} x t d t = x \int_{0}^{x} t d t = x \cdot \frac{x^{2}}{2} = \frac{x^{3}}{2} .

Now differentiate the resulting function of $x$ :

\frac{d}{d x} (\frac{x^{3}}{2}) = \frac{3 x^{2}}{2} .

The faulty answer $x^{2}$ misses the contribution of the $x$ that sits inside the integrand, which is a genuine second source of $x$ -dependence.

The Basic Rule

Combining with the Chain Rule

When Both Limits Are Variable

An Application to Indeterminate Limits

Implicit Differentiation of Integrals

Exercises

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