Derivative of this function?

[PandaherO]

1. Let the function f(x) have the property that f′(x)=x+1/x−3. If g(x)=f(x^2) find g′(x).

I've tried some steps already, however my answer is still wrong..

g'(x)=?
g'(x)=f'(x) at x^2 so, f'(x^2)?

(x+1)'(x-3)-(x+1)(x-3)'/(x-3)^2

in the end i get -4/(x-3)^2 and then I plug in x^2..
>-4/(x^4-6x^2+9)??
this seems to be wrong so could someone point out where my concept is flawed?
Thanks

 

[HallsofIvy]

I have no idea why you are finding the second derivative of f. That is not at all relevant to the question. Use the chain rule: $g(x)= f(x^2)$ so $g'(x)= f'(x^2)(x^2)'$. You are given f', not f, so you do not need to do that derivative.

 

[PandaherO]

Thanks so much, I don't exactly know what I was doing, either...haha I should have read the question more clearly.

 

[Ray Vickson]

1. Let the function f(x) have the property that f′(x)=x+1/x−3. If g(x)=f(x^2) find g′(x).

I've tried some steps already, however my answer is still wrong..

g'(x)=?
g'(x)=f'(x) at x^2 so, f'(x^2)?

(x+1)'(x-3)-(x+1)(x-3)'/(x-3)^2

in the end i get -4/(x-3)^2 and then I plug in x^2..
>-4/(x^4-6x^2+9)??
this seems to be wrong so could someone point out where my concept is flawed?
Thanks

Since g(x) = f(x^2), we have g'(x) = (d/dx) f(x^2), and you can write this out using the chain rule together with your formula f'(x) = x-3 + 1/x (as you have written).

RGV