How do you find the derivative of #1/x^2 #?

2 Answers
May 16, 2016

#-2/x^3#

Explanation:

We will use the power rule, which states that the derivative of #x^n# is #nx^(n-1)#.

We can use the power rule once we write #1/x^2# as #x^-2#.

Thus, according to the power rule, the derivative of #x^-2# is #-2x^(-2-1)=-2x^-3=-2/x^3#.

May 16, 2016

Use the limit definition to find:

#d/(dx) 1/x^2 = -2/x^3#

Explanation:

The power rule is good, but let's find it using the limit definition:

Let #f(x) = 1/x^2#

Then:

#d/(dx) f(x) = lim_(h->0) (f(x+h)-f(x))/h#

#= lim_(h->0)(1/(x+h)^2-1/x^2)/h#

#= lim_(h->0)(x^2-(x+h)^2)/(h(x+h)^2x^2)#

#= lim_(h->0)(x^2-(x^2+2hx+h^2))/(h(x+h)^2x^2)#

#= lim_(h->0)(-color(red)(cancel(color(black)(h)))(2x+h))/(color(red)(cancel(color(black)(h)))(x+h)^2x^2)#

#= lim_(h->0)(-(2x+h))/((x+h)^2x^2)#

#= (-2x)/x^4#

#= -2/x^3#