Question

Let $y=$$\sqrt{x}+2x^{\frac{3}{4}}+3x^{\frac{5}{6}}(x>0)$. Find the derivative of $y$ with respect to $x$.

Answer 1

Given,

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

Now,

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

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