Question

How do I find dy/dx of the following?

Answer 1

Answer:

$\displaystyle\frac{dy}{dx} \ = \ 3x^{2} \ + \ \displaystyle\frac{1}{2\sqrt{x^{3}}} \ - \ \displaystyle\frac{12}{x^{5}}$

Step-by-step explanation:

                      $y \ = \ x^{3} \ - \ \displaystyle\frac{1}{\sqrt{x}} \ + \ \displaystyle\frac{3}{x^{4}} \\ \\ y \ = \ x^{3} \ - \ x^{-\frac{1}{2}} \ + \ 3x^{-4} \\ \\ \displaystyle\frac{dy}{dx} \ = \ 3x^{3 \ - \ 1} \ - \ \left(-\displaystyle\frac{1}{2}\right)x^{-\frac{1}{2} \ - \ 1} \ + \ \left(-4 \ \times \ 3\right)x^{-4-1}$

                       $\displaystyle\frac{dy}{dx} \ = \ 3x^{2} \ + \ \displaystyle\frac{1}{2}x^{-\frac{3}{2}} \ - \ 12x^{-5} \\ \\ \displaystyle\frac{dy}{dx} \ = \ 3x^{2} \ + \ \displaystyle\frac{1}{2\sqrt{x^{3}}} \ - \ \displaystyle\frac{12}{x^{5}}$