find f′′(x). f(x)=(x² + 8)⁷ f′′(x)=□

Explanation:

Step1: Find first - derivative using chain - rule

Let $u = x^{2}+8$, then $y = u^{7}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=7u^{6}=7(x^{2}+8)^{6}$ and $\frac{du}{dx}=2x$. So, $f^{\prime}(x)=14x(x^{2}+8)^{6}$.

Step2: Find second - derivative using product - rule

The product - rule states that if $y = uv$, where $u = 14x$ and $v=(x^{2}+8)^{6}$, then $y^{\prime}=u^{\prime}v + uv^{\prime}$. $u^{\prime}=14$. To find $v^{\prime}$, use the chain - rule again. Let $t=x^{2}+8$, then $v = t^{6}$. $\frac{dv}{dt}=6t^{5}=6(x^{2}+8)^{5}$ and $\frac{dt}{dx}=2x$. So, $v^{\prime}=12x(x^{2}+8)^{5}$. $y^{\prime}=14(x^{2}+8)^{6}+14x\cdot12x(x^{2}+8)^{5}$.

Step3: Simplify the second - derivative

Factor out $14(x^{2}+8)^{5}$: $y^{\prime}=14(x^{2}+8)^{5}[(x^{2}+8)+12x^{2}]=14(x^{2}+8)^{5}(13x^{2}+8)$.

Answer:

$14(13x^{2}+8)(x^{2}+8)^{5}$