Find `(dy)/(dx),` `y=e^(3x)sin4x. 2^x`

Explanation

Let `y=e^{3 x} sin 4 x 2^{x}`
Take log both sides:
`Rightarrow log y=log (e^{3 x} sin 4 x 2^{x}) `
`Rightarrow log y=log (e^{3 x})+log (sin 4 x)+log (2^{x}) `
`{log (a b)=log a+log b ; log (frac{a}{b})=log a-log b} `
`Rightarrow log y=3 x log e+log (sin 4 x)+x log 2{log x^{a}=a log x} `
`Rightarrow log y=3 x+log (sin 4 x)+x log 2{log e=1}`
Differentiating with respect to `{x}` :
`Rightarrow frac{d(log y)}{dx}=frac{d(3 x)}{dx}+frac{d(log sin 4 x)}{dx}+frac{d(x log 2)}{dx}`
{Using chain rule, `frac{d(u+a)}{dx}=frac{du}{dx}+frac{da}{dx}` where `a` and `u` are any variables`Rightarrow frac{1}{y} frac{d y}{d x}=3 frac{d x}{d x}+frac{1}{sin 4 x} frac{d(sin 4 x)}{d x}+log 2 frac{d x}{d x}``frac{d(au)}{dx}=a frac{du}{dx} text { where } a text { is any constant and } u text { is any variable ; } frac{d(log u)}{dx}=frac{1}{u} frac{du}{dx}}`
`Rightarrow frac{1}{y} frac{d y}{d x}=3+frac{cos 4 x}{sin 4 x} frac{d(4 x)}{d x}+log 2 `
`{frac{d(sin u)}{d x}=cos u frac{d u}{d x}} `
`Rightarrow frac{1}{y} frac{d y}{d x}=3+cot 4 x times 4 frac{d x}{d x}+log 2 `
`Rightarrow frac{d y}{d x}=y{3+4 cot 4 x+log 2}`

Put the value of `y=e^{3 x} sin 4 x 2^{x} :`
`Rightarrow frac{d y}{d x}=e^{3 x} sin 4 x 2^{x}{3+4 cot 4 x+log 2}`