Question Number 83500 by 20092104 last updated on 03/Mar/20
$$\frac{{d}}{{dx}}\left(\left({a}+{bx}\right)^{{c}+{dx}} \right) \\ $$
Answered by john santu last updated on 03/Mar/20
![let y = (a+bx)^(c+dx) ln y = (c+dx) ln (a+bx) ((y′)/y) = d ln(a+bx) + ((b(c+dx))/(a+bx)) y′ = (dy/dx) = (a+bx)^(c+dx) [ d ln(a+bx) + ((b(c+dx))/(a+bx)) ]](https://www.tinkutara.com/question/Q83515.png)
$$\mathrm{let}\:\mathrm{y}\:=\:\left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{c}+\mathrm{dx}} \\ $$$$\mathrm{ln}\:\mathrm{y}\:=\:\left(\mathrm{c}+\mathrm{dx}\right)\:\mathrm{ln}\:\left(\mathrm{a}+\mathrm{bx}\right) \\ $$$$\frac{\mathrm{y}'}{\mathrm{y}}\:=\:\mathrm{d}\:\mathrm{ln}\left(\mathrm{a}+\mathrm{bx}\right)\:+\:\frac{\mathrm{b}\left(\mathrm{c}+\mathrm{dx}\right)}{\mathrm{a}+\mathrm{bx}} \\ $$$$\mathrm{y}'\:=\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{c}+\mathrm{dx}} \:\left[\:\mathrm{d}\:\mathrm{ln}\left(\mathrm{a}+\mathrm{bx}\right)\right. \\ $$$$\left.+\:\frac{\mathrm{b}\left(\mathrm{c}+\mathrm{dx}\right)}{\mathrm{a}+\mathrm{bx}}\:\right]\: \\ $$