Text Compute the derivatives of the following composite functions: (a) [esin(x)]′\big[e^{\sin(x)}\big]'[esin(x)]′ (b) [ln(x2+2x)]′\big[\ln(x^2+2x)\big]'[ln(x2+2x)]′ (c) [cos(x3)]′\big[\cos(x^3)\big]'[cos(x3)]′ (d) [ln(x)]′\big[\sqrt{\ln(x)}\big]'[ln(x)]′ Solution Chain rule: [f(g(x))]′=f′(g(x))⋅g′(x)\big[f(g(x))\big]' = f'(g(x))\cdot g'(x)[f(g(x))]′=f′(g(x))⋅g′(x). (a) [esinx]′=esinx⋅cosx=cosx esinx\big[e^{\sin x}\big]' = e^{\sin x}\cdot\cos x = \cos x\,e^{\sin x}[esinx]′=esinx⋅cosx=cosxesinx (b) [ln(x2+2x)]′=2x+2x2+2x\big[\ln(x^2+2x)\big]' = \dfrac{2x+2}{x^2+2x}[ln(x2+2x)]′=x2+2x2x+2 (c) [cos(x3)]′=−sin(x3)⋅3x2=−3x2sin(x3)\big[\cos(x^3)\big]' = -\sin(x^3)\cdot 3x^2 = -3x^2\sin(x^3)[cos(x3)]′=−sin(x3)⋅3x2=−3x2sin(x3) (d) [lnx]′=12lnx⋅1x=12xlnx\big[\sqrt{\ln x}\big]' = \dfrac{1}{2\sqrt{\ln x}}\cdot\dfrac{1}{x} = \dfrac{1}{2x\sqrt{\ln x}}[lnx]′=2lnx1⋅x1=2xlnx1 cosx esinx; 2x+2x2+2x; −3x2sin(x3); 12xlnx\boxed{\cos x\,e^{\sin x};\ \ \dfrac{2x+2}{x^2+2x};\ \ -3x^2\sin(x^3);\ \ \dfrac{1}{2x\sqrt{\ln x}}}cosxesinx; x2+2x2x+2; −3x2sin(x3); 2xlnx1