Logarithmic Differentiation — Question 4

Question 4 - Solution

For $x>1$, define the positive base

$$B(x)=\frac{x^2\sqrt{x+1}}{(x^2+4)^3\ln x},\qquad f(x)=B(x)^{\sin x}.$$

Set

$$A(x)=\ln B(x)=2\ln x+\tfrac12\ln(x+1)-3\ln(x^2+4)-\ln(\ln x).$$

Then $\ln f(x)=\sin x\,A(x)$ and

$$A'(x)=\frac2x+\frac1{2(x+1)}-\frac{6x}{x^2+4}-\frac1{x\ln x}.$$

Logarithmic differentiation yields

$$\boxed{f'(x)=B(x)^{\sin x}\bigl(\cos x\,A(x)+\sin x\,A'(x)\bigr),\quad x>1.}$$

Here $B,A,A'$ are the explicit expressions above.