Answer to Question #311349 in Calculus for Ankit

Question

Find the 𝑛th order differential coefficient of 𝑦 = sin 𝑥 log(𝑎𝑥 + 𝑏).

Answer 1

$y=sinx log(ax+b)$

Using product rule, we have

$\frac{dy}{dx}(UV)=U\frac{dv}{dx}+V\frac{du}{dx}$

$\frac{dy}{dx}=sinx\frac{d}{dx}log(ax+b)+log(ax+b)\frac{d}{dx}sinx$

$$=sinx(\frac{1\times a}{ax+b})+log(ax+b)cosx$$

$\frac{d^2y}{dx}=$ $sinx\frac{d}{dx}(\frac{1\times a}{ax+b})+(\frac{1\times a}{ax+b})\frac{d}{dx}sinx+ log(ax+b)cosx+cos\frac{d}{dx}log(ax+b)$

$$=sinx\frac{(-1)^1\times a^2}{(ax+b)^2}+\frac{1\times a}{ax+b}cosx+log(ax+b)(-sinx) +cosx\frac{(-1)^1\times a^2}{(ax+b)^2}$$

$\frac{d^3y}{dx}=sinx\frac{(-1)^2\times a^3\times2}{(ax+b)^3} +\frac{(-1)^1\times a^2}{(ax+b)^2}cosx+\frac{1\times a}{ax+b}(-sinx) +cosx\frac{(-1)^1\times a^2}{(ax+b)^2}+log(ax+b)(-cosx) +cosx\frac{(-1)^2\times a^3\times 2}{(ax+b)^3}$

$$=sinx\frac{-1^2\times a^3\times 2}{(ax+b)^2}+2(cosx\frac{-1^1\times a^2}{(ax+b)^2} +log(ax+b)(-cosx)+cosx\frac{(-1)^2\times a^3\times 2}{(ax+b)^3}$$

$\frac{d^4y}{dx}=sinx\frac{-1^3\times a^4\times 6}{(ax+b)^4} +\frac{-1^2\times a^3\times 2}{(ax+b)^3}cosx +\frac{-1^1\times a^2}{(ax+b)^2}(-sinx)+\frac{-1^2\times a^3\times 2}{(ax+b)^3}cosx +\frac{1\times a}{ax+b}(-cosx)+(-sinx)\frac{-1^1\times a^2}{(ax+b)^4} +cosx\frac{-1^2\times a^3\times 2}{(ax+b)^3} +\frac{-1^1\times a^2}{(ax+b)^2}(-sinx)+ log(ax+b)sinx+(-cosx)\frac{1\times a}{ax+b}+cosx\frac{-1^3\times a^4\times 6}{(ax+b)^4} +(-sinx)\frac{-1^2\times a^3\times 2}{(ax+b)^3}$

$=sinx\frac{-1^3\times a^4\times 6}{(ax+b)^4}+3(cosx\frac{-1^2\times a^3\times 2}{(ax+b)^3}) -3(sinx\frac{-1^1\times a^2}{(ax+b)^2})-2(cosx\frac{1\times a}{ax+b}) +log(ax+b)sinx -sinx\frac{-1^2\times a^3\times 2}{(ax+b)^3}+cosx\frac{-1^3\times a^4\times 6}{(ax+b)^4}$

$\frac{d^ny}{dx}=$ $sinx\frac{-1^{n-1}(n-1){!}a^n}{(ax+b)^{n}})+(n-1)(cos x\frac{-1^{n-2}(n-2){!}a^{n-1}}{(ax+b)^{n-1}}) -(n-1)(sinx\frac{-1^{n-3}(n-3){!}a^{n-2}}{(ax+b)^{n-2}}) -(n-2)(cos x\frac{1\times a}{ax+b}) +log(ax+b)(sin(x+\frac{n\pi}{2})-(sinx\frac{-1^{n-2}(n-2){!}a^{n-1}}{(ax+b)^{n-1}})+cos x\frac{-1^{n-1}(n-1){!}a^n}{(ax+b)^{n}})$