MHB Finding F'(x) for $\int_{1}^{2x}\frac{1}{1-t^{3}} \,dx$

[karush]

if $f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dx$ dt, then $F'(x)=$

is this just $\frac{-(2x-1)}{t^2-1}$ ?

 

[I like Serena]

if $f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dx$ dt, then $F'(x)=$

is this just $\frac{-(2x-1)}{t^2-1}$ ?

Huh?

Do you perhaps mean:
$$f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dt$$
and
$$f'(x)=?$$

 

[karush]

Do you perhaps mean:
$$f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dt$$
and
$$f'(x)=?$$

yes

 

[I like Serena]

yes

Ah okay. Then the result should contain only $x$ and no $t$.To figure this out let's define $g(t)=\frac{1}{1-t^{3}}$.
And let's suppose we know $G(t)$, so that $G'(t)=g(t)$.

Then:
$$f(x)=\int_{1}^{2x}\frac{1}{1-t^{3}} \,dt = \int_{1}^{2x}g(t) \,dt = G(2x) - G(1)$$
We'll leave $G$ like that, since we don't really know $G$.

Taking the derivative, we get:
$$f'(x) = G'(2x) \cdot 2 - 0 = g(2x)\cdot 2 = \frac{1}{1-(2x)^{3}} \cdot 2$$

Hey! We got rid of $G$, so we have the answer! :D

 

[Dethrone]

Not my question, but I want to say that was a very good excellent explanation. (Smoking)

 

[karush]

wow that's a cool way to do it(Cool)