Question

If y + (sinx)^tanx, then dy/dx is equal to

(A) (sinx)^tanx . (1 + sec²x . logsinx)

(B) tanx. (sinx)^{tanx - 1} . cosx

(C) (sinx)^tanx . sec²x . log sinx

(D) tanx . (sinx)^{tanx - 1}

Answer 1

Answer is (A) (sinx)^tanx . (1 + sec²x . logsinx)

Given y = (sinx)^tanx; logy = tanx . logsinx

Differentiating with respect to x, we get

1/y .dy/dx = tanx . cotx + logsinx . sec²x

dy/dx = (sinx)^tanx [1 + logsinx. sec²x]