Question

find the derivative of the function.
y = cos(sqrt(sin(tan(3x))))
y =

Explanation:

Step1: Apply chain - rule outer function

The outer function is $y = \cos(u)$ where $u=\sqrt{\sin(\tan(3x))}$. The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$. So we have $y'=-\sin(\sqrt{\sin(\tan(3x))})\cdot\frac{d}{dx}(\sqrt{\sin(\tan(3x))})$.

Step2: Apply chain - rule to square - root function

Let $v = \sin(\tan(3x))$, then $\sqrt{v}=v^{\frac{1}{2}}$. The derivative of $v^{\frac{1}{2}}$ with respect to $v$ is $\frac{1}{2}v^{-\frac{1}{2}}$. So $\frac{d}{dx}(\sqrt{\sin(\tan(3x))})=\frac{1}{2\sqrt{\sin(\tan(3x))}}\cdot\frac{d}{dx}(\sin(\tan(3x)))$.

Step3: Apply chain - rule to sine function

Let $w=\tan(3x)$. The derivative of $\sin(w)$ with respect to $w$ is $\cos(w)$. So $\frac{d}{dx}(\sin(\tan(3x)))=\cos(\tan(3x))\cdot\frac{d}{dx}(\tan(3x))$.

Step4: Apply chain - rule to tangent function

The derivative of $\tan(3x)$ with respect to $3x$ is $\sec^{2}(3x)$, and the derivative of $3x$ with respect to $x$ is $3$. So $\frac{d}{dx}(\tan(3x)) = 3\sec^{2}(3x)$.

Step5: Combine all results

$y'=-\sin(\sqrt{\sin(\tan(3x))})\cdot\frac{1}{2\sqrt{\sin(\tan(3x))}}\cdot\cos(\tan(3x))\cdot3\sec^{2}(3x)=-\frac{3\sec^{2}(3x)\cos(\tan(3x))\sin(\sqrt{\sin(\tan(3x))})}{2\sqrt{\sin(\tan(3x))}}$.

Answer:

$-\frac{3\sec^{2}(3x)\cos(\tan(3x))\sin(\sqrt{\sin(\tan(3x))})}{2\sqrt{\sin(\tan(3x))}}$