Question

If $x^4+4^x+a^x+x^a=1$, find $\frac{dy}{dx}$.

Answer 1

Let $y=x^4+4^x+a^x+x^a-1$

$\frac{dy}{dx}=\frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(4^x\right)+\frac{d}{dx}\left(a^x\right)+\frac{d}{dx}\left(x^a\right)+10$

$\frac{dy}{dx}=4x^3+4^x\log 4+a^x\log a+a{x}^{a-1}$

$[\because \frac{d}{dx}{x}^n=n{x}^{n-1}.\frac{d}{dx}{a}^x=a^x\log a]$