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OpenStudy (anonymous):

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 ago
OpenStudy (mathteacher1729):

Do you know the quotient rule?

7 years ago
OpenStudy (anonymous):

yes

7 years ago
OpenStudy (mathteacher1729):

You 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 ago
OpenStudy (anonymous):

thank you :)

7 years ago
OpenStudy (anonymous):

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

7 years ago
OpenStudy (anonymous):

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

7 years ago
OpenStudy (anonymous):

its x^2 not x^-2

7 years ago
OpenStudy (anonymous):

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.

7 years ago
OpenStudy (anonymous):

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

7 years ago
OpenStudy (anonymous):

ah ok thanks

7 years ago
OpenStudy (anonymous):

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

7 years ago
OpenStudy (anonymous):

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

7 years ago
OpenStudy (anonymous):

well 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
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