Question

question 17
find the derivative of $y = e^{5x}\cos(3x)$.
\\(\frac{dy}{dx} = -15\sin(3x)\\)
\\(\frac{dy}{dx} = -3e^{5x}\sin(3x)\\)
\\(\frac{dy}{dx} = 5e^{5x}\cos(3x) - 3e^{5x}\sin(3x)\\)
\\(\frac{dy}{dx} = 5e^{5x}\cos(3x) + 3e^{5x}\sin(3x)\\)

Explanation:

Step1: Recall product rule

If $y = u(x)v(x)$, then $\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)$. Let $u(x)=e^{5x}$, $v(x)=\cos(3x)$.

Step2: Compute $u'(x)$

Derivative of $e^{5x}$: $u'(x)=5e^{5x}$

Step3: Compute $v'(x)$

Derivative of $\cos(3x)$: $v'(x)=-3\sin(3x)$

Step4: Apply product rule

$\frac{dy}{dx}=5e^{5x}\cos(3x) + e^{5x}(-3\sin(3x))$
Simplify: $\frac{dy}{dx}=5e^{5x}\cos(3x)-3e^{5x}\sin(3x)$

Answer:

$\boldsymbol{\frac{dy}{dx}=5e^{5x}\cos(3x)-3e^{5x}\sin(3x)}$ (corresponding to the third option)