How do you use the definition of a derivative to find the derivative of 4/x^2?

2 Answers
Mar 2, 2016

Just remember that 4/(x^2)=4x^(-2) and that (x^n)'=n x^(n-1).

So (4/(x^2))'=(4x^(-2))'= 4xx(-2)x^(-3)=-8/x^3

Mar 3, 2016

See the explanation section below.

Explanation:

Let f(x) = 4/x^2

I will assume that you are using the limit definition in the form

f'(x) = lim_(hrarr0) (f(x+h)-f(x))/h

So, we get:

f'(x) = lim_(hrarr0)(4/(x+h)^2-4/x^2)/h " " (indeterminate form 0/0)

= lim_(hrarr0)((4x^2- (4(x+h)^2))/(x^2((x+h)^2)))/(h/1)

= lim_(hrarr0)(4x^2-(4x^2+8xh+4h^2))/(x^2(x+h)^2)*1/h

= lim_(hrarr0)(-8xh-4h^2)/(x^2(x+h)^2)*1/h

= lim_(hrarr0)(-8x-4h)/(x^2(x+h)^2) " " (not an indeterminate form)

= (-8x)/x^4 = (-8)/x^3

If you are using f'(x) = lim_(trarrx)(f(x)-f(t))/(x-t)

Then the algebra is similar.

(f(x)-f(t))/(x-t) = (4/x^2-4/t^2)/(x-t)

= ((4t^2-4x^2)/(x^2t^2))/((x-t)/1)

= (4(t^2-x^2))/(x^2t^2)*1/(x-t)

= (-4(x^2-t^2))/(x^2t^2)*1/(x-t)

= (-4(x+t)(x-t))/(x^2t^2)*1/(x-t)

= (-4(x+t))/(x^2t^2)

Taking the limit as trarrx, we get

(-4(2x))/(x^2x^2) = (-8)/x^3