Pregunta: Find \frac{dy}{dx} wheny=-8\frac{1}{\sin^{4}\!\left(\ln\!\left(x\right)\right)}\frac{dy}{dx}=

To find $\left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)$ when $(\mathrm{y}=8{\csc }^4\left(\ln \left(\mathrm{x}\right)\right)),$you'll need to use the chain rule.

Let's denote$(\mathrm{u}=\ln \left(\mathrm{x}\right))$, then $\left(\mathrm{y}\right)$ can be expressed as $(\mathrm{y}=8{\csc }^4\left(\mathrm{u}\right)).$