What is the value of f'(2) when f(x) and x are given in a polynomial equation?

[Macroer]

Homework Statement

[f(x)]^5 + f(x)/(7x^2)=4
when f(2)=3, calculate f'(2)

Homework Equations

The Attempt at a Solution

Attachment has solution

I am stuck. I don't know if I am right upto there.

 

[Dick]

I can't see your attachment. But just take the derivative of both sides of the equation and solve for f'(x).

 

[HallsofIvy]

Here is what is in the attachment:

$$\left[f(x)\right]^5+ f(x)/7x^2= 4$$
$$7x^2\left[f(x)\right]^5+ f(x)= 28x^2$$
$$f(x)= 28x^2- 7x^2\left[f(x)\right]^5$$
$$f'(x)= 56x- (14x\left[f(x)\right]^5- 35x^2\left[f(x)\right]^4 f'(x))$$

That has a sign wrong- either drop the parentheses or write
$$f'(x)= 56x- (14x\left[f(x)\right]^5+ 35x^2\left[f(x)\right]^4 f'(x))$$

when x= 2,
$$f'(2)= 56(2)- (14(2)3^6+ 35(4)(3)^4f'(x))$$
$$f'(2)= -6692- 11340f'(2)$$

and you now have the sign right!

Now, just solve for f'(2): add 11340f'(2) to both sides and divide by 11341.
(I didn't check your arithmetic!)