Question

differenciate x^n logx​

Answer 1

ANSWER

d(x^n logx) = x^n d(logx) + logx d(x^n)

dx dx dx

= x^n.1/x + logx . n ( x) ^(n-1)

= x^(n-1) + n.logx. x^(n-1)

= x^(n-1) [ 1+ n logx]

Formula used

d(u.v)/dx = v. du/dx + u. dv/dx

Standered results

d(x^n)/dx= n x^(n-1)

d(logx) /dx = 1/x

hope it helps