Question

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

Answer 1

Given,

$y=x^4+4^x+\log x-1$

Differentiate on both sides w.r.t x

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

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

$\therefore \frac{dy}{dx}=4x^3+\frac{1}{x}+\left(\log 4\right)4^x$