# Derivative of an Integral problem

#### AlderDragon

I need to find the derivative of an integral, with a limit of integration being a function:

$$\displaystyle g(x) = \int_1^{x^{4}}(t^3+2)dt$$

How do I find g'(x)?

#### CaptainBlack

MHF Hall of Fame
I need to find the derivative of an integral, with a limit of integration being a function:

$$\displaystyle g(x) = \int_1^{x^{4}}(t^3+2)dt$$

How do I find g'(x)?
Put:

$$\displaystyle h(x)=\int_1^{x}(t^3+2)dt$$

Then:

$$\displaystyle g(x)=h(x^4)$$

and so by the chain rule:

$$\displaystyle \frac{d}{dx}g(x)=4x^3 \left. \frac{dh}{dx}\right|_{x^4}$$

where the derivative of $$\displaystyle h$$ is obtained by applying the fundamental theorem of calculus.

CB

AlderDragon

#### AlderDragon

Upon closer inspection of my past work I found another method of doing this type of problem. Substitute $$\displaystyle x^4$$ in for $$\displaystyle t$$ and multiply by the derivative of $$\displaystyle x^4$$, $$\displaystyle 4x^3$$:

$$\displaystyle g(x) = \int_1^{x^{4}}(t^3+2)dt$$

$$\displaystyle g'(x) = (x^{12} + 2)4x^3$$

Is this the correct answer? And, if so, is this a different answer than you might obtain using a different method?

#### CaptainBlack

MHF Hall of Fame
Upon closer inspection of my past work I found another method of doing this type of problem. Substitute $$\displaystyle x^4$$ in for $$\displaystyle t$$ and multiply by the derivative of $$\displaystyle x^4$$, $$\displaystyle 4x^3$$:

$$\displaystyle g(x) = \int_1^{x^{4}}(t^3+2)dt$$

$$\displaystyle g'(x) = (x^{12} + 2)4x^3$$

Is this the correct answer? And, if so, is this a different answer than you might obtain using a different method?
That is what you get if you finish what I posted in my last post, so yes it is correct. However why you do what you say you do (and what it means for that matter) needs explaining.

CB

#### HallsofIvy

MHF Helper
By the way, here is the general Leibniz formula:
$$\displaystyle \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} F(x,t)dt=$$$$\displaystyle F(x, \beta(x))\frac{d\beta(x)}{dx}- F(x,\alpha(t))\frac{d\alpha(x)}{dx}+$$$$\displaystyle \int_{\alpha(x)}^{\beta(x)}$$$$\displaystyle \frac{\partial F(x,t)}{\partial x} dt$$.

It can be proved from the fundamental theorem of calculus and using the chain rule for both the upper and lower limits.

AlderDragon