Question

If f(x)=x^2+1 and g(x)=2x-1, find f'(g(x)) at x=1.

Answer 1

hello f ( g(x) ) = (g(x))^2 + 1 its a good beginning

Answer 2

So it's (2x-1)^2+1?

Answer 3

yes now you must develop the operations inside

Answer 4

then you should find f´

Answer 5

Since f'(x)=2x, then 2(4x^2-4x+2)=8x^2-8x+4, plug in 1 gives you 4, am I right?

Answer 6

\[\huge f(x)=x^2+1 \qquad f'(x)=2x\]
\[\large f(g(x))=(2x-1)^2+1 \qquad \rightarrow \qquad f(g(x))=4x^2-4x\]\[\large f'(g(x))=8x-4\]\[\large f'(g(1))=8-4 \qquad \rightarrow \qquad f'(g(1))=4\]

Answer 7

Hmm I'm a little confused by the way you did that, but it looks like it turned out ok :)

Answer 8

Thanks for your help.