Question Number 6233 by sanusihammed last updated on 19/Jun/16
$${Diffrentiate}\:\:\:\:{y}\:\:=\:\:{x}^{\mathrm{3}{ln}\mathrm{5}} \\ $$
Answered by malwaan last updated on 19/Jun/16
$${y}^{'} =\mathrm{3}{ln}\mathrm{5}{x}^{\mathrm{3}{ln}\mathrm{5}−\mathrm{1}} \\ $$
Answered by uchechukwu okorie favour last updated on 10/Aug/16
$${add}\:{natural}\:{log}\:{to}\:{both}\:{sides} \\ $$$${In}\mid{y}\mid={In}\mid{x}^{\mathrm{3}{In}\mathrm{5}} \mid \\ $$$${In}\mid{y}\mid=\mathrm{3}{In}\mathrm{5}{Inx} \\ $$$${differentiate}\:{both}\:{sides} \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}=\mathrm{3}{In}\mathrm{5}.\frac{\mathrm{1}}{{x}} \\ $$$${where}\:\mathrm{3}{In}\mathrm{5}\backsimeq\mathrm{4}.\mathrm{8283}\:\left(\mathrm{4}{d}.{p}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}=\frac{\mathrm{4}.\mathrm{8283}}{{x}} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{4}.\mathrm{8283}}{{x}}×{y} \\ $$$${where}\:{y}={x}^{\mathrm{3}{In}\mathrm{5}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{4}.\mathrm{8283}}{{x}}\boldsymbol{\div}{x}^{\mathrm{4}.\mathrm{8283}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}.\mathrm{8283}{x}^{\mathrm{4}.\mathrm{8283}−\mathrm{1}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}.\mathrm{8283}{x}^{\mathrm{3}.\mathrm{8283}} \:{or}\:\mathrm{3}{In}\mathrm{5}{x}^{\mathrm{3}{In}\mathrm{5}−\mathrm{1}} \\ $$$$ \\ $$