Question

I have the answer for this and will put in the comments. I just don't understand the steps:
Find the derivative of g(x) Integral with range of (1,2x) (1)/(t^3+1) dt with respect to x.

Answer 1

The explanation is: If we let y=2x then the derivative of g(x) = integral (1,y) of (1)/(t^3 + 1) dt with respect to y will be 1/y^3+1 from the Fundamental Theorem; its derivative with respect to x we can then find using the Chain Rule to be g ' (x) = 2/(2x)^3+1. ~~~~~ It's that last part that I don't get....the chain rule to derive 1/y^3+1???

Answer 2

You initially took the derivative with respect to y because you substituted y=2x to deal with that limit not being simply one variable.
But because you took it with respect to y and we wanted it with respect to x, we have an additional step of getting back our x:
\( \dfrac{d}{dx} (Integral) = \color{blue}{\dfrac{dy}{dx}} \dfrac{d}{dy} (Integral)|_{y=2x} \)
The derivative with respect to y, then, you back-substitute y =2x to get that back in terms of x.

Answer 3

And \( \dfrac{1}{y^2 + 1} \) is our result from taking the derivative with respect to y. So we multiply the derivative of y with respect to x: y=2x, dy/dx = 2. We also substitute 2x back in for y.

Does that resolve your uncertainty or did I miss that nail? :)

Answer 4

\( \dfrac{1}{y^3 + 1} \), not y^2. Sorry.

Answer 5

Sorry for still being confused, but I am.
Let me start from the beginning. To attack this problem I need to first substitute y for the 2x in the limit. So I have:
g(x)=\[\int\limits\limits\limits_{1}^{y} \frac{ 1 }{ t^3+1 } dt\]
Then I am supposed to take the derivative of the \[\frac{ 1 }{ t^3+1 }\] correct?

Answer 6

At that point, we are using this theorem:
\( \displaystyle \dfrac{d}{dx} \int_{a}^{x} f(t) \ dt = f(x) \)
In our problem, we changed that limit into just a y, so this only applies when the derivative is taken to the same variable as that limit. Like this:
\( \displaystyle \dfrac{d}{dy} \int_{a}^{y} f(t) \ dt = f(y) \)

Answer 7

But the problem is that the original question doesn't want the derivative in terms of y, it wants the derivative in terms of x. So when we do this:
\( \quad \displaystyle \frac{d}{d\color{red}{x}} \int_{a}^{\color{green}y} f(t) \ dt \)
We use chain rule to change it into the derivative with respect to y, at the expense of an extra factor to balance it out (the dy's would "cancel" because one is in the numerator and one the denominator).
\( \quad \displaystyle \frac{d\color{green}y}{d\color{red}x} \frac{d}{d\color{green}{y}} \int_{a}^{\color{Green}{y}} f(t) \ dt \)