Question

worked example: derivatives of sin(x) and cos(x)
related content
ab
unit 2

d sin(x), >

\frac{d}{dx}3cos(x)+4x^{2}=square

Explanation:

Step1: Apply sum - rule of differentiation

The derivative of a sum $u + v$ is $\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}$. Let $u = 3\cos(x)$ and $v = 4x^{2}$.

Step2: Differentiate $u = 3\cos(x)$

The derivative of $\cos(x)$ is $-\sin(x)$. Using the constant - multiple rule $\frac{d}{dx}(cf(x))=c\frac{d}{dx}(f(x))$, where $c = 3$ and $f(x)=\cos(x)$, we get $\frac{d}{dx}(3\cos(x))=3\frac{d}{dx}(\cos(x))=- 3\sin(x)$.

Step3: Differentiate $v = 4x^{2}$

Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, where $n = 2$ and $c = 4$, we have $\frac{d}{dx}(4x^{2})=4\times2x=8x$.

Step4: Combine the results

$\frac{d}{dx}(3\cos(x)+4x^{2})=\frac{d}{dx}(3\cos(x))+\frac{d}{dx}(4x^{2})=-3\sin(x)+8x$.

Answer:

$-3\sin(x)+8x$