Question

Find the derivative
$\sqrt{x}+2.x^{3/4}+3.x^{5/6}$

Answer 1

Consider the given function

$y=\sqrt{x}+2.x^{\frac{3}{4}}+3.x^{\frac{5}{6}}$

Differentiating this equation with respect to x, and we get,

$\frac{dy}{dx}=\frac{d}{dx}(\sqrt{x}+2.x^{\frac{3}{4}}+3.x^{\frac{5}{6}})$

$=\frac{1}{2\sqrt{x}}+2\times \frac{3}{4}{x}^{\frac{3}{4}-1}+3\times \frac{5}{6}{x}^{\frac{5}{6}-1}$

$=\frac{1}{2\sqrt{x}}+\frac{3}{2}{x}^{-\frac{1}{4}}+\frac{5}{2}{x}^{\frac{-1}{6}}$

$=\frac{1}{2\sqrt{x}}+\frac{3}{2x^{\frac{1}{4}}}+\frac{5}{2x^{\frac{1}{6}}}$

Hence, this is the answer.