find f(x). f(x) = (x² + 4)⁵

Explanation:

Step1: Find the first - derivative using the chain rule

Let $u = x^{2}+4$, then $y = u^{5}$. The chain rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=5u^{4}=5(x^{2}+4)^{4}$ and $\frac{du}{dx}=2x$. So, $f^{\prime}(x)=5(x^{2}+4)^{4}\cdot2x = 10x(x^{2}+4)^{4}$.

Step2: Find the second - derivative using the product rule

The product rule states that if $y = uv$, where $u = 10x$ and $v=(x^{2}+4)^{4}$, then $y^{\prime}=u^{\prime}v + uv^{\prime}$. $u^{\prime}=10$. To find $v^{\prime}$, use the chain rule again. Let $t=x^{2}+4$, then $v = t^{4}$. $\frac{dv}{dt}=4t^{3}=4(x^{2}+4)^{3}$ and $\frac{dt}{dx}=2x$, so $v^{\prime}=4(x^{2}+4)^{3}\cdot2x = 8x(x^{2}+4)^{3}$. $y^{\prime}=10(x^{2}+4)^{4}+10x\cdot8x(x^{2}+4)^{3}$.

Step3: Simplify the second - derivative

Factor out $10(x^{2}+4)^{3}$: $y^{\prime}=10(x^{2}+4)^{3}[(x^{2}+4)+8x^{2}]$. $y^{\prime}=10(x^{2}+4)^{3}(9x^{2}+4)$.

Answer:

$10(x^{2}+4)^{3}(9x^{2}+4)$