Question

If $y=x^x+x^7+7^x+7^7$, then $\frac{dy}{dx}=?$

Answer 1

$y=x^x+x^7+7^x+7^{7}Let{x}^x=u$

Taking log both the sides,

$\log x^x=\log u=>x\log x=\log u$

Differentiating both sides with respect to 'x',

$\frac{1}{u}\frac{du}{dx}=\frac{x}{x}+\log x=>\frac{du}{dx}=u\times (1+\log x)=>\frac{du}{dx}=x^x\times (1+\log x)$

Now, $y=x^x+x^7+7^x+7^7=>y=u+x^7+7^x+7^7=>\frac{dy}{dx}=\frac{du}{dx}+7x^6+7^x\log 7+0=>\frac{dy}{dx}=x^x\times (1+\log x)+7x^6+7^x\log 7$