Question

Differentiat w. r. t. x : ${x^{x}}^{3}+(x^{3})^{x}$

Answer 1

Answer:

$\bf\frac{dy}{dx}=x^{x^3+2}[1+3\:logx]+3(x^3)^x[1+logx]$

Step-by-step explanation:

$\text{Let }y=x^{x^3}+(x^3)^x=u+v\:(say)$

$\implies\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$u=x^{x^3}$

$\text{Take logarithm on both sides }$

$log\:u=log\:x^{x^3}$

$log\:u=x^3\:log\:x$

$\text{differentiate with respect to x}$

$\frac{1}{u}\frac{du}{dx}=x^3(\frac{1}{x})+logx\:3x^2$

$\frac{du}{dx}=u[x^2+logx\:3x^2]$

$\frac{du}{dx}=x^{x^3}[x^2+logx\:3x^2]$

$\frac{du}{dx}=x^{x^3}x^2[1+3\:logx]$

$\bf\frac{du}{dx}=x^{x^3+2}[1+3\:logx]$

$v=(x^3)^x$

$\text{Take logarithm on both sides }$

$log\:v=x\:log\:x^3$

$log\:v=3\:xlogx$

$\text{differentiate with respect to x}$

$\frac{1}{v}\frac{dv}{dx}=3[x(\frac{1}{x})+logx.1]$

$\frac{dv}{dx}=3v[1+logx]$

$\frac{dv}{dx}=3(x^3)^x[1+logx]$

$\bf\frac{dv}{dx}=3(x^3)^x[1+logx]$

$\therefore\:\bf\frac{dy}{dx}=x^{x^3+2}[1+3\:logx]+3(x^3)^x[1+logx]$