calculate the derivative of the following function. y = cos^5(sin(3x)) dy/dx = □

Explanation:

Step1: Apply chain - rule for outer function

Let $u = \sin(3x)$. Then $y=\cos^{5}(u)$. The derivative of $y$ with respect to $u$ using the power - rule and the derivative of cosine is $\frac{dy}{du}=5\cos^{4}(u)\times(-\sin(u))=- 5\cos^{4}(u)\sin(u)$.

Step2: Apply chain - rule for inner function

The derivative of $u = \sin(3x)$ with respect to $x$ is $\frac{du}{dx}=\cos(3x)\times3 = 3\cos(3x)$.

Step3: Use the chain - rule formula

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$. Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$ into the formula: [ \begin{align*} \frac{dy}{dx}&=-5\cos^{4}(\sin(3x))\sin(\sin(3x))\times3\cos(3x)\ &=-15\cos(3x)\cos^{4}(\sin(3x))\sin(\sin(3x)) \end{align*} ]

Answer:

$-15\cos(3x)\cos^{4}(\sin(3x))\sin(\sin(3x))$