proove derivative of x^-n =-n x^ (-n-1)
derivative proof for x^-n
2 Answers
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.
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)


