Question

the derivative of the following function ( f(x)=2 - e^{-2x}cos(5x + 2)) simplifies to ( f(x)=g(x)e^{-2x}), where ( g(x)) is a function with no factors of ( e^{-2x}). find ( g(x)). ( g(x)=) symbolic expression

Explanation:

Step1: Apply product - rule and chain - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, $u = 2 - e^{-2x}$ and $v=\cos(5x + 2)$. First, find the derivative of $u$ and $v$. The derivative of $u$ with respect to $x$: $u'=\frac{d}{dx}(2 - e^{-2x})=0-(- 2)e^{-2x}=2e^{-2x}$ (using the chain - rule $\frac{d}{dx}(e^{ax})=ae^{ax}$). The derivative of $v$ with respect to $x$: $v'=\frac{d}{dx}(\cos(5x + 2))=-5\sin(5x + 2)$ (using the chain - rule $\frac{d}{dx}(\cos(ax + b))=-a\sin(ax + b)$).

Step2: Calculate $f'(x)$

By the product - rule $f'(x)=u'v+uv'$. Substitute $u$, $u'$, $v$, and $v'$ into the formula: $f'(x)=2e^{-2x}\cos(5x + 2)+(2 - e^{-2x})(-5\sin(5x + 2))$. Expand the expression: $f'(x)=2e^{-2x}\cos(5x + 2)-10\sin(5x + 2)+5e^{-2x}\sin(5x + 2)$. Factor out $e^{-2x}$: $f'(x)=e^{-2x}(2\cos(5x + 2)+5\sin(5x + 2))-10\sin(5x + 2)$. We want $f'(x)=g(x)e^{-2x}$, so $g(x)=2\cos(5x + 2)+5\sin(5x + 2)$.

Answer:

$g(x)=2\cos(5x + 2)+5\sin(5x + 2)$