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

7 years agoDo you know the quotient rule?

7 years agoyes

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

7 years agothank you :)

7 years agois the answer 0 (or i am getting it wrong?)

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

7 years agoits x^2 not x^-2

7 years agoI 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.

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

7 years agoah ok thanks

7 years agoi am just wondering why is that different from my answer using quotient rule

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

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

7 years ago