Question

given the function y = -3 sin(3√x), find dy/dx.

Explanation:

Step1: Apply chain - rule

Let $u = 3\sqrt{x}=3x^{\frac{1}{2}}$, then $y=-3\sin(u)$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$.
$\frac{dy}{du}=-3\cos(u)$

Step2: Differentiate $u$ with respect to $x$

Differentiate $u = 3x^{\frac{1}{2}}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $\frac{du}{dx}=3\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{3}{2\sqrt{x}}$.

Step3: Calculate $\frac{dy}{dx}$

Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$ into the chain - rule formula: $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=-3\cos(u)\cdot\frac{3}{2\sqrt{x}}$. Replace $u = 3\sqrt{x}$ back into the expression.
$\frac{dy}{dx}=-\frac{9\cos(3\sqrt{x})}{2\sqrt{x}}$

Answer:

$-\frac{9\cos(3\sqrt{x})}{2\sqrt{x}}$