Derivative of an integral
So you’ve mastered derivatives (NBD) — but now you need to find the derivative of an integral? What even is an integral?
We can answer all that and more!
Let’s start here: An integral is a set of all the antiderivatives of the function. So a derivative of an integral is a derivative of an antiderivative of a function.
Did that just blow your mind a little? Honestly, same. Let’s see what all that really means!
What does it mean to find the derivative of an integral?
The integral is an antiderivative, and differentiation (or finding a derivative) is the inverse procedure of integration (or finding the integral).
As we mentioned, finding the derivative of an integral means finding the derivative of the antiderivative, which is defined by the second fundamental theorem of calculus.
The second fundamental theorem of calculus states that if $$f$$ is continuous on $$[a,b]$$ and $$a\leq x\leq b$$, the derivative of the integral of $$f$$ can be calculated as follows:
Don’t forget to keep our differentiation rules close by — we still need them!
| Constant multiple property of derivatives | $$\frac{d}{dx}\left(c\times f(x)\right)=c\times\frac{d}{dx}\left(f(x) \right)$$ |
| Sum rule for derivatives | $$\frac{d}{dx}\left(f(x) + g(x)\right)=\frac{d}{dx}\left( f(x) \right)+\frac{d}{dx}\left( g(x) \right)$$ |
| Difference rule for derivatives | $$\frac{d}{dx}\left(f(x) - g(x)\right)=\frac{d}{dx}\left( f(x) \right)-\frac{d}{dx}\left( g(x) \right)$$ |
| Product rule for derivatives | $$\frac{d}{dx}\left(f(x)\times g(x)\right)=\frac{d}{dx}\left( f(x) \right)\times g(x)+f(x)\times\frac{d}{dx}\left( g(x) \right) $$ |
| Quotient rule for derivatives | $$\frac{d}{dx}\left(\frac{f(x)}{g(x)} \right)=\frac{\frac{d}{dx}\left(f(x) \right)\times g(x)-f(x)\times\frac{d}{dx}\left( g(x) \right)}{(g(x))^{2}}$$ |
| The Chain rule | $$(f\circ g)^{\prime}(x)=f^{\prime}(g(x))\times g^{\prime}(x) \text{ or } \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} \text{ when } y=f(u),u=g(x)$$ |
| Derivative of the inverse function | $$\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}\left(x\right)\right)}$$ |
Why is the derivative of an integral so useful?
You’ve already learned how to solve linear, quadratic, and all sorts of other equations. But before you learned to solve them, you learned the basic calculations that you needed, like adding, subtracting, multiplying, etc.
Finding the derivative of an integral is kind of like basic arithmetic for solving differential equations, which you’ll learn really soon. So, by mastering these skills now, you’re setting yourself up for success down the road!
How to find the derivative of an integral
Now that we know a little more about derivatives of integrals, let’s walk through a few problems together.
Example 1
Find the derivative of an integral:
To find the derivative, apply the second fundamental theorem of calculus, which states that if $$f$$ is continuous on $$[a,b]$$ and $$a\leq x \leq b$$, the derivative of an integral of $$f$$ can be calculated $$\frac{d}{dx} \int_{a}^{x} f(t)dt=f(x)$$:
$$x^5$$
So, the derivative of an integral $$\frac{d}{dx}\int_{0}^{x} t^5dt$$ is:
Example 2
Find the derivative of an integral:
Substitute $$u$$ for $$x^3$$:
$$\frac{d}{dx}\int_{\frac{\pi}{2}}^{u} \cos(t)dt$$
We’ll use the chain rule to find the derivative, because we want to transform the integral into a form that works with the second fundamental theorem of calculus:
$$\frac{d}{du} \left( \int_{\frac{\pi}{2}}^{u} \cos\left(t\right)dt \right) \times\frac{du}{dx}$$
Nice! Now we can find the derivative by using the second fundamental theorem of calculus, which states that if $$f$$ is continuous on $$[a,b]$$ and $$a\leq x \leq b$$, the derivative of an integral of $$f$$ can be calculated $$\frac{d}{dx} \int_{a}^{x} f(t)dt=f(x)$$:
$$\cos{u}\times \frac{du}{dx}$$
Rewrite the derivative:
$$\cos{u}\times \frac{d}{dx}(u)$$
Substitute back $$u=x^3$$. We can do this because the substituted values don’t represent the actual solution (they just simplify the calculation of one part of the problem!).
$$\cos{x^3}\times \frac{d}{dx}(x^3)$$
Find the derivative of $$x^3$$ by using the differentiation rule $$\frac{d}{dx} x^n=nx^{n-1}$$:
$$\cos{x^3}\times 3x^2$$
Use the commutative property to reorder the terms:
$$3x^2 \times \cos{x^3}$$
Here we go! The derivative of an integral $$\frac{d}{dx}\int_{\frac{\pi}{2}}^{x^3} \cos(t)dt$$ is:
That wasn’t so bad, right? When you start working on other problems, remember these steps and you’ll be golden:
Study summary
- Rewrite the integral as a sum so that only one limit of integration in both integrals depends on the independent variable.
- Use the chain rule to find the derivative.
- Apply the second fundamental theorem of calculus.
- If the substitution was used, substitute back.
- Find the derivative of the expression.
Do it yourself!
Ready to keep honing your skills? Try applying those steps to these practice problems!
Take the derivative of a function:
- $$\frac{d}{dx}\int_{\pi}^{x^2} t^3dt$$
- $$\frac{d}{dx}\int_{1}^{x^2} \sqrt{5t}dt$$
- $$\frac{d}{dx}\int_{x}^{\frac{\pi}4} (t^2-\ln(t))^2dt$$
- $$\frac{d}{dx}\int_{-x}^{x} \ln{(2+\sin{t})}dt$$
Solutions:
- $$2x^7$$
- $$2\sqrt5x\times|x|$$
- $$-x^4+2x^2+\ln{x}-\ln(x)^2$$
- $$\ln(2-\sin{x})+\ln{(2+\sin{x})}$$
If you’re struggling through the solving process, that’s totally okay! Stumbling a few times is actually good for learning. If you can’t find your way out of the problem, scan it with your Photomath app, and we’ll walk you through to the other side!
Here’s a sneak peek of what you’ll see:
