derivative proof for x^-n

Question

proove derivative of x^-n =-n x^ (-n-1) 

Answer 1

Definition of derivative: lim h->0 (f(x + h) –f(x))/h

((x + h)^-n - x^-n)/h

= (x^n - (x + h)^n)/(x + h)^n*(x)^n*(h)

use the binomial expansion theorem, and we are interested in only the first 3 terms.

= (x^n - (x^n + nx^(n – 1)h + (n(n-1))/2)x^(n-2)h^2 +…………))/((x + h)^n(x)^n(h))

The first two terms in the numerator become 0, the third term becomes -nx^(n – 1), the h cancels out, and the remaining numerator terms vanish as h-> 0.

We are left with:

= -nx^(n – 1)/ (x + h)^n*(x)^n

Take the limit as h ->0 and

= -nx^(n – 1)/ x^2n

= -nx^(-n-1)

Obviously the below solution is more elegant, but going back to the definition of a derivative will always work.

Answer 2

first let y = x^-n.

Now use logarithmic properties.

ln y = ln x^-n.

ln y = -n ln x 

Now use logarithmic differentiation on both sides.

y'/y = -n 1/x

y'/y = -n/x   (mult. -n * 1/x)

Now sove for y'

y' = y * -n/x

Now substitute x^-n back in for y.

y' = x^-n *-n/x

y' = (x^-n * -n)/x          (mult. x^-n *n/x)

y' = -n * (x^-n)/x

y' = -n * x^-n-1       (by exponent rules)