find dy.\ny = sin(15\\sqrt{x})\ndy = \\square dx

Explanation:

Step1: Let $u = 15\sqrt{x}=15x^{\frac{1}{2}}$

$y=\sin(u)$

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

$\frac{dy}{du}=\cos(u)$

Step3: Differentiate $u$ with respect to $x$

$\frac{du}{dx}=15\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{15}{2\sqrt{x}}$

Step4: Use the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

$\frac{dy}{dx}=\cos(15\sqrt{x})\cdot\frac{15}{2\sqrt{x}}$

Step5: Find $dy$

$dy = \frac{15\cos(15\sqrt{x})}{2\sqrt{x}}dx$

Answer:

$\frac{15\cos(15\sqrt{x})}{2\sqrt{x}}$