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#