Question from shambodeb, a student:

This is a successive differentiation problem by Leibnitz theorem

If y = x^{n-1} log x ; Proof nth derivative y^{(n)} = (n-1)!/x

Shambodeb,

I would prove this by induction.

It is straightforward to verify it is true for n = 1.

Inductive hypotheses: Suppose that for some positive integer k, if y = x^{k-1} log x then the k^{th} derivative of y is y^{(k)} = (k-1)!/x.

Let y = x^{k+1-1} log x = x^{k} log x and find the (k + 1)^{st} derivative of y.

y' = k x^{k-1} log x + x^{k} × (1/x) = k x^{k-1} log x + x^{k-1} Differentiate k more times using the induction hypothesis.

y' = k x^{k-1} log x + x^{k} × (1/x) = k x^{k-1} log x + x^{k-1}

Differentiate k more times using the induction hypothesis.

Harley