Find $f'(1)$ given $f(x)$ and $f(1)$

Question

Question:

Let $$f(x)=\int\frac{x-1}{x+1}\frac{dx}{\sqrt{x^3+x^2+x}}$$ such that $f(1)=\frac{2\pi}{3}$

Then $f'(1)$ is equal to

A) $0$

B) $\cfrac \pi3$

C) $\cfrac\pi4$

D) $2$

Attempt:

\begin{align}f(x)&=\int\frac{x-1}{x+1}\frac{1}{\sqrt{x^3+x^2+x}}dx\\ f(1)&=\int0dx\\ f(1)&=0+c\\ &= \frac{2\pi}{3}\\ \implies c&= \frac{2\pi}{3} \end{align}

I don't know what to do beyond this step. This is an indefinite integral so we cannot differentiate it. How else we bring $f'(x)$ into picture? Any hints?

Answer 1

The stipulation that $f(1)=2\pi/3$ is a red herring; it's irrelevant to the solution. As soon as you see a relation of the form $f(x)=\int g(x)dx$, you know that $f'(x)=g(x)$. In this case,

$$f'(x)={x-1\over (x+1)\sqrt{x^3+x^2+x}}$$

and thus $f'(1)=0$.

Answer 2

Does $f'(x)$ mean $\frac{df}{dx}$. Isnt $f'(x)$ then simply the integrated term, with the integration removed? ie, $\frac{x-1}{x+1}\times \frac{1}{\sqrt{x + x^2 + x^3}}$ ? Am I missing something?

Answer 3

$f'(x)$ is simply the term under integration.$$f'(x)=\frac{x-1}{x+1}\frac{1}{\sqrt{x + x^2 + x^3}}$$

Take an example.

$$\text{Let} ~~~g(x) = \int xdx ; ~~\text{find g(1), given g(0)=0.}$$.

As you have done , it should be $$g(1) = \int 1dx= x+c$$ using $ Since g(0)=0 \implies c=0$

So, $g(1)$ should $x$, isn't it weird?

Yes, it is, because it is wrong.

By integrating , we will get $$g(x)=\frac{x^2}{2} \implies g(1)=\frac12$$

Therefore, $$f'(1)=\frac{1-1}{1+1}\frac{1}{\sqrt{1 + 1^2 + 1^3}}=0$$

Remember, if :

$$f(x)=\int g(x)dx \implies f'(x)=g(x) ~; ~~$$