Question

$\frac{d}{dx}left(\frac{4(6.4)(-x + 4sqrt{3})}{sqrt{(-x + 4sqrt{3})^2+25}}+\frac{(-4sqrt{3})(32)}{sqrt{(-x + 4sqrt{3})^2+25}}
ight)$

Explanation:

Step1: Identify the form

This is a derivative of a sum of two rational - functions. Use the sum - rule of differentiation $\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}$.
Let $u=\frac{4(6.4)(-x + 4\sqrt{3})}{\sqrt{(-x + 4\sqrt{3})^2+25}}$ and $v=\frac{(-4\sqrt{3})(3z)}{\sqrt{(-x + 4\sqrt{3})^2+25}}$. Since $z$ is treated as a constant with respect to $x$, $\frac{dv}{dx}=0$.

Step2: Use the quotient - rule

The quotient - rule states that if $y=\frac{f(x)}{g(x)}$, then $y^\prime=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{g(x)^2}$.
For $u=\frac{f(x)}{g(x)}$ where $f(x)=4(6.4)(-x + 4\sqrt{3})=-25.6x+102.4\sqrt{3}$ and $g(x)=\sqrt{(-x + 4\sqrt{3})^2+25}=(x^{2}-8\sqrt{3}x + 73)^{\frac{1}{2}}$.
First, find $f^\prime(x)$: $f^\prime(x)=-25.6$.
Second, find $g^\prime(x)$ using the chain - rule. Let $u = x^{2}-8\sqrt{3}x + 73$, then $g(x)=u^{\frac{1}{2}}$. $\frac{du}{dx}=2x-8\sqrt{3}$ and $\frac{dg}{du}=\frac{1}{2}u^{-\frac{1}{2}}$. So $g^\prime(x)=\frac{2x - 8\sqrt{3}}{2\sqrt{(-x + 4\sqrt{3})^2+25}}=\frac{x - 4\sqrt{3}}{\sqrt{(-x + 4\sqrt{3})^2+25}}$.

Step3: Apply the quotient - rule formula

$\frac{du}{dx}=\frac{-25.6\sqrt{(-x + 4\sqrt{3})^2+25}-(-25.6x + 102.4\sqrt{3})\frac{x - 4\sqrt{3}}{\sqrt{(-x + 4\sqrt{3})^2+25}}}{(-x + 4\sqrt{3})^2+25}$.
Simplify the numerator:
\[

$$\begin{align*} &-25.6((-x + 4\sqrt{3})^2+25)-(-25.6x + 102.4\sqrt{3})(x - 4\sqrt{3})\\ =&-25.6(x^{2}-8\sqrt{3}x + 73)-(-25.6x^{2}+102.4\sqrt{3}x+102.4\sqrt{3}x - 1228.8)\\ =&-25.6x^{2}+204.8\sqrt{3}x-1884.8 + 25.6x^{2}-204.8\sqrt{3}x + 1228.8\\ =&- 656 \end{align*}$$

\]
So $\frac{du}{dx}=\frac{-656}{((-x + 4\sqrt{3})^2+25)^{\frac{3}{2}}}$ and $\frac{d}{dx}(u + v)=\frac{-656}{((-x + 4\sqrt{3})^2+25)^{\frac{3}{2}}}$.

Answer:

$\frac{-656}{((-x + 4\sqrt{3})^2+25)^{\frac{3}{2}}}$