YAMT: Derivative (Edit: Thanks TuxDave)

[Chaotic42]

I'm going back over some stuff, and here's the problem:

If f(4)=3 and f'(4)=-5, find g'(4):

g(x)=sqr(x)*f(x)

So I did this:

g'(x)=[sqr(x)*f(x)]'=[x^(-1/2) * f(x) + sqr(x) * f'(x)] : Product rule
g'(x)=[sqr(x)*f(x)]'=[1/(sqr(x)] * f(x) + sqr(x) * f'(x)]

So if x=4:

g'(4)=[1/2] * 3 + 2 * -5
g'(4)=1.5-10
g'(4)=-8.5

The book says the answer is -9.25

I know it's something stupid, but what am I doing wrong? I've gotten every other answer in the chapter review so far correct (this is #29).

Edit: I must be getting tired, I forgot to bring down the (1/2) as per the power rule. Thanks.

 

[TuxDave]

Originally posted by: Chaotic42

So I did this:

g'(x)=[sqr(x)*f(x)]'=[x^(-1/2) * f(x) + sqr(x) * f'(x) 😛roduct rule

Math error

should be
0.5x^-0.5*f(x) + sqrt(x)*f'(x)

so plugging in the above

0.5*1/2*3+2*-5 = 10-3/4 = -9.25

 

[Chaotic42]

Originally posted by: TuxDave

Math error

should be
0.5x^-0.5*f(x) + sqrt(x)*f'(x)

so plugging in the above

0.5*1/2*3+2*-5 = 10-3/4 = -9.25

Edit: Ah, I see. Thanks.

 

[TuxDave]

Originally posted by: Chaotic42

So I did this:

g'(x)=[sqr(x)*f(x)]'=[x^(-1/2) * f(x) + sqr(x) * f'(x) 😛roduct rule

Math error

should be
0.5x^-0.5*f(x) + sqrt(x)*f'(x)