supposed that f(1)=1 and f'(1)=3. What is the value of the derivative of g(x)=(f(x)/x^2) evaluated at the point x=1

Question

mathteacher1729 · Answer

Do you know the quotient rule?

anonymous · Answer

yes

mathteacher1729 · Answer

You want to use the quotient rule here, but wherever you see "f(x)" and need to take its derivative, just write "f ' (x)" . :)

anonymous · Answer

thank you :)

anonymous · Answer

is the answer 0 (or i am getting it wrong?)

anonymous · Answer

g(x)=f(x)*x^-2
g'(x)=f'(x)*x^-2+f(x)*-2x^-3
           =3*1+1*-2*1
          =1

anonymous · Answer

its x^2 not x^-2

anonymous · Answer

I think you wrote that it was divided by x^2 which is the same as multiplying by x^-2 so you can use the product rule rather than the quotieint rule which I think is eaisier.

anonymous · Answer

and why do you use product rule instead of quotient rule? becasue g(x)= f(x)/x^2

anonymous · Answer

ah ok
thanks

anonymous · Answer

i am just wondering why is that different from my answer using quotient rule

anonymous · Answer

sorry, i made mistake, its 1 as well :) sorry...

anonymous · Answer

well it should be the same. 
g'(x)=(f'(x)*x^2-f(x)2x)/x^4
   =(3*1-1*2*1)/1
=1