Question

calculate the derivative of the following function.
y = sin(3x^5 + 7x + 3)
\frac{dy}{dx}=

Explanation:

Step1: Identify outer - inner functions

Let $u = 3x^{5}+7x + 3$, then $y=\sin(u)$.

Step2: Differentiate outer function

The derivative of $y = \sin(u)$ with respect to $u$ is $\frac{dy}{du}=\cos(u)$.

Step3: Differentiate inner function

The derivative of $u = 3x^{5}+7x + 3$ with respect to $x$ is $\frac{du}{dx}=15x^{4}+7$.

Step4: Apply chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=\cos(u)$ and $\frac{du}{dx}=15x^{4}+7$ back in, and replace $u$ with $3x^{5}+7x + 3$. So $\frac{dy}{dx}=\cos(3x^{5}+7x + 3)\cdot(15x^{4}+7)$.

Answer:

$(15x^{4}+7)\cos(3x^{5}+7x + 3)$