Question

question
find $\frac{dy}{dx}$ if $y = (-3x^{5}-3)^{3}$.
provide your answer below:
$\frac{dy}{dx}=square$

Explanation:

Step1: Apply chain - rule for differentiation

Let \(u=-3x^{5}-3\), so \(y = u^{3}\). First find \(\frac{dy}{du}\) and \(\frac{du}{dx}\).
The derivative of \(y = u^{3}\) with respect to \(u\) is \(\frac{dy}{du}=3u^{2}\).
The derivative of \(u=-3x^{5}-3\) with respect to \(x\) is \(\frac{du}{dx}=-15x^{4}\).
By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), so \(\frac{dy}{dx}=3(-3x^{5}-3)^{2}\cdot(- 15x^{4})=-45x^{4}(-3x^{5}-3)^{2}\).

Step2: Find \(\frac{dx}{dy}\)

Since \(\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}\), then \(\frac{dx}{dy}=-\frac{1}{45x^{4}(-3x^{5}-3)^{2}}\)

Answer:

\(-\frac{1}{45x^{4}(-3x^{5}-3)^{2}}\)