+0

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

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1153
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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

Oct 17, 2017
edited by Guest  Oct 17, 2017
edited by Guest  Oct 17, 2017

#1
+7348
+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. )

Oct 18, 2017
#2
+17747
+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.

Oct 18, 2017