Answer
$\frac{dy}{dx} = 4(\frac{x^{2}}{8} + x - \frac{1}{x})^{3} \times (\frac{x}{4} +1 + \frac{1}{x^{2}})$
Work Step by Step
$y = u^{4}$
where,
$u = (\frac{x^{2}}{8} + x - \frac{1}{x})$
Now,
$\frac{dy}{du} = 4u^{3}$
and,
$\frac{du}{dx} = (\frac{x}{4} +1 + \frac{1}{x^{2}})$
So,
$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$
$\frac{dy}{dx} = 4u^{3} \times (\frac{x}{4} +1 + \frac{1}{x^{2}})$
$\frac{dy}{dx} = 4(\frac{x^{2}}{8} + x - \frac{1}{x})^{3} \times (\frac{x}{4} +1 + \frac{1}{x^{2}})$
