Question

calculate the derivative of $h(x)=2sin x + 3cos x$. select the correct answer below: 2cos x + 3sin x -2cos x + 3sin x -2cos x - 3sin x 2cos x - 3sin x

Explanation:

Step1: Use sum - rule of derivatives

The sum - rule states that if $h(x)=f(x)+g(x)$, then $h'(x)=f'(x)+g'(x)$. Here $f(x) = 2\sin x$ and $g(x)=3\cos x$, so $h'(x)=(2\sin x)'+(3\cos x)'$.

Step2: Apply constant - multiple rule

The constant - multiple rule is $(cf(x))'=cf'(x)$. So, $(2\sin x)' = 2(\sin x)'$ and $(3\cos x)'=3(\cos x)'$.

Step3: Recall derivatives of sine and cosine

We know that $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$. Then $2(\sin x)' = 2\cos x$ and $3(\cos x)'=- 3\sin x$.

Step4: Combine the results

$h'(x)=2\cos x-3\sin x$.

Answer:

$2\cos x - 3\sin x$ (corresponding to the last option)