+0

# If f(x)= then f'(2)= ?

0
190
2

If $$f(x)= [x+1/x-1]^3$$ then f'(2)= ?

$$f'(x)= 3[x+1/x-1]^2$$just not sure what to do after that

Guest Oct 17, 2017
edited by Guest  Oct 17, 2017
edited by Guest  Oct 17, 2017
Sort:

#1
+6335
+1

Remember...

$$\frac{d}{dx}u^3 \,=\, 3u^2\,\cdot\,\frac{d}{dx}u$$

So

$$f'(x)\,=\,3[x+\frac1x-1]^2\,\cdot\,\frac{d}{dx}[x+\frac1x-1] \\~\\ f'(x)\,=\,3[x+\frac1x-1]^2\,\cdot\,[\frac{d}{dx}x+\frac{d}{dx}\frac1x-\frac{d}{dx}1] \\~\\ f'(x)\,=\,3[x+\frac1x-1]^2\,\cdot\,[1-x^{-2}]$$

To find  f'(2) , plug in  2  for  x .

$$f'(2)\,=\,3[2+\frac12-1]^2\,\cdot\,[1-2^{-2}] \\~\\ f'(2)\,=\,3[\frac32]^2\,\cdot\,[1-\frac1{4}] \\~\\ f'(2)\,=\,3[\frac94]\,\cdot\,[\frac3{4}] \\~\\ f'(2)\,=\,\frac{81}{16}$$

( If   f(x)  is meant to be  $$[\frac{x+1}{x-1}]^3$$  then you get a different answer. )

hectictar  Oct 18, 2017
#2
+17711
+2

If  f(x)  =  [x + 1/x - 1]3,  f(x)  is of the form  f(x) = [g(x)]3, so  f'(x)  =  3[g(x)]2·g'(x)

f(x)  =  [x + 1/x - 1]3     --->     f'(x)  =  3[x + 1/x - 1]3·[1 - 1/x2]

To find f'(2), replace x with 2 in the above derivative, and simplify.

geno3141  Oct 18, 2017

### 23 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details