Question

Help! logarithmic differentiation

Answer 1

Recall that we can write

`f(x)^(g(x))=e^{\ln f(x)^(g(x))}=e^(g(x)\ln f(x))`

so that when we take the derivative, we get by the chain rule

`\left(f(x)^(g(x))\right)'=(g(x)\ln f(x))' e^(g(x)\ln f(x))=(g(x)\ln f(x))'f(x)^(g(x))`

Here, we have
`f(x)=3-\sin2x` and
`g(x)=\sqrt[3]{x}=x^(1/3)`. By the product rule,

`\left(x^(1/3)\ln(3-\sin2x)\right)'=\left(x^(1/3)\right)\ln(3-\sin2x)+x^(1/3)(\ln(3-\sin2x))'`

`=\frac13x^(-2/3)\ln(3-\sin2x)+x^(1/3)((3-\sin2x)')/(3-\sin2x)`

`=\frac13x^(-2/3)\ln(3-\sin2x)+x^(1/3)(2\cos2x)/(\sin2x-3)`

`=\frac1{3x^(2/3)}\left(\ln(3-\sin2x)+3x(2\cos2x)/(\sin2x-3)\right)`

`=\frac1{3\sqrt[3]{x^2}}\left(\ln(3-\sin2x)+(6x\cos2x)/(\sin2x-3)\right)`

So we have

`y'=\frac{(3-\sin2x)^{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}\left(\ln(3-\sin2x)+(6x\cos2x)/(\sin2x-3)\right)`