Show solutionHide solution+ Question 4 For x>1x>1 (so the base is positive), let the function be: f(x)=(x2⋅x+1(x2+4)3⋅lnx)sinxf(x) = \left( \frac{x^2 \cdot \sqrt{x + 1}}{(x^2 + 4)^3 \cdot \ln x} \right)^{\sin x} Use logarithmic differentiation to find f′(x)f'(x). Simplify your answer as much as possible. Question 4 - Solution For x>1x>1, define the positive base B(x)=x2x+1(x2+4)3lnx,f(x)=B(x)sinx.B(x)=\frac{x^2\sqrt{x+1}}{(x^2+4)^3\ln x},\qquad f(x)=B(x)^{\sin x}. Set A(x)=lnB(x)=2lnx+12ln(x+1)−3ln(x2+4)−ln(lnx).A(x)=\ln B(x)=2\ln x+\tfrac12\ln(x+1)-3\ln(x^2+4)-\ln(\ln x). Then lnf(x)=sinxA(x)\ln f(x)=\sin x\,A(x) and A′(x)=2x+12(x+1)−6xx2+4−1xlnx.A'(x)=\frac2x+\frac1{2(x+1)}-\frac{6x}{x^2+4}-\frac1{x\ln x}. Logarithmic differentiation yields f′(x)=B(x)sinx(cosxA(x)+sinxA′(x)),x>1.\boxed{f'(x)=B(x)^{\sin x}\bigl(\cos x\,A(x)+\sin x\,A'(x)\bigr),\quad x>1.} Here B,A,A′B,A,A' are the explicit expressions above.