Question

find $\frac{dy}{dx}$ c) $f(x)=(x + 2)^6(2x^2+7)^4$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Let $u=(x + 2)^6$ and $v=(2x^2+7)^4$.

Step2: Find $\frac{du}{dx}$

Using the chain - rule, if $u=(x + 2)^6$, let $t=x + 2$, then $u=t^6$. $\frac{du}{dt}=6t^5$ and $\frac{dt}{dx}=1$. So, $\frac{du}{dx}=6(x + 2)^5$.

Step3: Find $\frac{dv}{dx}$

Using the chain - rule, if $v=(2x^2+7)^4$, let $s = 2x^2+7$, then $v=s^4$. $\frac{dv}{ds}=4s^3$ and $\frac{ds}{dx}=4x$. So, $\frac{dv}{dx}=4(2x^2+7)^3\cdot4x=16x(2x^2+7)^3$.

Step4: Calculate $\frac{dy}{dx}$

$\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}=(x + 2)^6\cdot16x(2x^2+7)^3+(2x^2+7)^4\cdot6(x + 2)^5$.
Factor out $2(x + 2)^5(2x^2+7)^3$:
\[

$$\begin{align*} \frac{dy}{dx}&=2(x + 2)^5(2x^2+7)^3[8x(x + 2)+3(2x^2+7)]\\ &=2(x + 2)^5(2x^2+7)^3(8x^2+16x + 6x^2+21)\\ &=2(x + 2)^5(2x^2+7)^3(14x^2+16x + 21) \end{align*}$$

\]

Answer:

$2(x + 2)^5(2x^2+7)^3(14x^2+16x + 21)$