Question

Find ${\displaystyle \frac{dy}{dx}ify=x^3–2x^2+\sqrt{x}+1}$

Answer 1

${\displaystyle y=x^3–2x^2+\sqrt{x}+1}$
Differentiating w.r.t. x, we get

${\displaystyle \frac{dy}{dx}=\frac{d}{dx}(x^3-2x^2+\sqrt{x}+1)}$

=${\displaystyle \frac{d}{dx}\left(x^3\right)-2\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\sqrt{x}\right)+\frac{d}{dx}\left(1\right)}$

= ${\displaystyle 3x^2-2\left(2x\right)+\frac{d}{dx}\left(x^{\frac{1}{2}}\right)+0}$

=${\displaystyle 3x^2-4x+\frac{1}{2}{x}^{\frac{1}{2}-1}}$

=${\displaystyle 3x^2-4x+\frac{1}{2}{x}^{\frac{-1}{2}}}$

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