Differentiable equations

[idk90]

Let g(x)=x^(2)*sqrt(f(x)) where f is a differentiable function such that f(1)=4 and f'(1)=1. Find g'(1). The answer if 17/4.

so far I solved for g'(x) which I believe comes to 2x*sqrt(f(x))+x^2*d/dx*f(x)/2*sqrt(f(x))

and then I tried to sub in the given. so I subbed 1 into every x I see but can't seem to come to the final answer?

 

[Romsek]

I answered this question today at one of these forums.

Anddd it looks like you got that answer.

 

[Deleted member 4993]

Let g(x)=x^(2)*sqrt(f(x)) where f is a differentiable function such that f(1)=4 and f'(1)=1. Find g'(1). The answer if 17/4.

so far I solved for g'(x) which I believe comes to 2x*sqrt(f(x))+x^2*d/dx*f(x)/2*sqrt(f(x))

and then I tried to sub in the given. so I subbed 1 into every x I see but can't seem to come to the final answer?

Please share your work , step-by-step , so that we can catch the mistake ...

 

[Steven G]

so far I solved for g'(x) which I believe comes to 2x*sqrt(f(x))+x^2*d/dx*f(x)/2*sqrt(f(x)).

That is NOT true what you wrote above. What does d/dx* mean? Are you computing the derivative of * or perhaps *f(x)/2*sqrt(f(x))? Neither makes any sense! It should be g'(x) = 2x*sqrt(f(x))+x^2*f '(x)/2*sqrt(f(x))