Question

Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^3 3.sin(xe^x)^3 4.sin^3(xe^x) 5.ln(xy)=e^2y

Answer 1

(1) I assume "log" on its own refers to the base-10 logarithm.

`\left(e^(-2x)\log(\ln x)^3\right)'=\left(e^(-2x)\right)'\log(\ln x)^3+e^(-2x)\left(\log(\ln x)^3\right)'`

`=-2e^(-2x)\log(\ln x)^3+(e^(-2x))/(\ln10(\ln x)^3)\left((\ln x)^3\right)'`

`=-2e^(-2x)\log(\ln x)^3+(3e^(-2x)(\ln x)^2)/(\ln10(\ln x)^3)\left(\ln x\right)'`

`=-2e^(-2x)\log(\ln x)^3+(3e^(-2x)(\ln x)^2)/(\ln10\,x(\ln x)^3)`

`=-2e^(-2x)\log(\ln x)^3+(3e^(-2x))/(\ln10\,x\ln x)`

Note that writing
`\log(\ln x)^3=3\log(\ln x)` is one way to avoid using the power rule.

(2)

`\left(e^(-2x)(\log(\ln x))^3\right)'=(e^(-2x))'(\log(\ln x))^3+e^(-2x)\left(\log(\ln x))^3\right)'`

`=-2e^(-2x)(\log(\ln x))^3+3e^(-2x)(\log(\ln x))^2(\log(\ln x))'`

`=-2e^(-2x)(\log(\ln x))^3+3e^(-2x)(\log(\ln x))^2((\ln x)')/(\ln10\,\ln x)`

`=-2e^(-2x)(\log(\ln x))^3+(3e^(-2x)(\log(\ln x))^2)/(\ln10\,x\ln x)`

(3)

`\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^(3x))\right)'=\cos(x^3e^(3x)(x^3e^(3x))'`

`=\cos(x^3e^(3x))((x^3)'e^(3x)+x^3(e^(3x))')`

`=\cos(x^3e^(3x))(3x^2e^(3x)+3x^3e^(3x))`

`=3x^2e^(3x)(1+x)\cos(x^3e^(3x))`

(4)

`\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'`

`=3\sin^2(xe^x)\cos(xe^x)(xe^x)'`

`=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')`

`=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)`

`=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)`

(5) Use implicit differentiation here.

`(\ln(xy))'=(e^(2y))'`

`((xy)')/(xy)=2e^(2y)y'`

`(x'y+xy')/(xy)=2e^(2y)y'`

`y+xy'=2xye^(2y)y'`

`y=(2xye^(2y)-x)y'`

`y'=\frac y{2xye^(2y)-x}`