Question

if f(x)=(-2x^2 + 2x + 7)cos(x), find f(x)=

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=-2x^{2}+2x + 7$ and $v(x)=\cos(x)$.

Step2: Find $u'(x)$

Differentiate $u(x)=-2x^{2}+2x + 7$ using the power - rule. The power - rule is $\frac{d}{dx}(ax^{n})=nax^{n - 1}$. So, $u'(x)=-4x + 2$.

Step3: Find $v'(x)$

Differentiate $v(x)=\cos(x)$. We know that $\frac{d}{dx}(\cos(x))=-\sin(x)$.

Step4: Calculate $f'(x)$

Using the product - rule $f'(x)=u'(x)v(x)+u(x)v'(x)$. Substitute $u(x),u'(x),v(x),v'(x)$:
\[

$$\begin{align*} f'(x)&=(-4x + 2)\cos(x)+(-2x^{2}+2x + 7)(-\sin(x))\\ &=(-4x + 2)\cos(x)+(2x^{2}-2x - 7)\sin(x) \end{align*}$$

\]

Answer:

$(-4x + 2)\cos(x)+(2x^{2}-2x - 7)\sin(x)$