Applying the chain rule with the fundamental theorem of calculus 1
I have the following problem in which I have to apply both the chain rule and the FTC 1. I got the right answer, but i'm confused about what's really going.
$$\frac{d}{dx} \int_1^{x^4} sec(t) \space dt $$
While trying to make sense of this, I have the following in mind.
If $g(x)=\int_a^{x}f(t)\space dt$
then $g(x)^{'}=f(x)$
And $(f(g(x)){'}$= $f^{'}(g(x)) \times g^{'}(x)$
and
$\frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx}$
Applying these to the this integral, i'm trying to break everything down to see what is what. So that for example I know which function is nested in which function.
I'm just looking for a clear and simple explanation of what's going on here.
$\endgroup$1 Answer
$\begingroup$To avoid some of the confusion and to see how the Chain Rule applies here, use some other names of functions when stating the Chain Rule. What you wrote as the Chain Rule is perfectly correct and this is how it's stated in probably any textbook out there… but it makes you confused because in your question $g$ is the outside function, not inside, and $f$ is a totally different thing (not the outside function of the Chain Rule).
So let's do some renaming. If we call the outside function $g(x)$ and the inside function $h(x)$, then the very same Chain Rule will be written as $$\left[g(h(x))\right]'=g'(h(x))\cdot h'(x).$$ In your example: $$g(x)=\int_a^x f(t)\,dt, \quad \text{so} \quad g'(x)=f(x);$$ and $$h(x)=x^4, \quad \text{so} \quad h'(x)=4x^3.$$ The function $h(x)=x^4$ is the inside function here because it replaces $x$ in the expression for the outside function $g(x)$.
$\endgroup$ 1
