# How do you use the quotient rule to differentiate (1+3x)^2 / (1+sqrtx)^2?

Jan 14, 2018

$\frac{9 {x}^{2} \sqrt{x} + 27 {x}^{2} + 18 x \sqrt{x} + 6 x + 5 \sqrt{x} - 1}{\sqrt{x} {\left(\sqrt{x} + 1\right)}^{4}}$

#### Explanation:

The quotient rule states that,

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$

Here, $u = {\left(3 x + 1\right)}^{2}$, $v = {\left(\sqrt{x} + 1\right)}^{2}$

So, we have to differentiate $u$ using the chain rule, which says that

d/dx(f(g(x))=color(blue)(f'(g(x)))*color(red)(g'(x))

Here, the function is ${\left(3 x + 1\right)}^{2}$, the inside derivative is $3$, so

$\frac{d}{\mathrm{dx}} \left({\left(3 x + 1\right)}^{2}\right) = \textcolor{b l u e}{2 \left(3 x + 1\right)} \cdot \textcolor{red}{3} = 6 \left(3 x + 1\right)$

With the same method, we can differentiate $v$, the function is ${\left(\sqrt{x} + 1\right)}^{2}$, inside derivative is $\frac{1}{2 \sqrt{x}}$.

So,

d/dx((sqrt(x)+1)^2)=color(blue)(2(sqrt(x)+1)*color(red)(1/(2sqrt(x)))=color(black)((sqrt(x)+1)/sqrt(x))

Now, we put those back into the quotient rule,

$\frac{d}{\mathrm{dx}} \left({\left(3 x + 1\right)}^{2} / {\left(\sqrt{x} + 1\right)}^{2}\right) = \frac{6 \left(3 x + 1\right) {\left(\sqrt{x} + 1\right)}^{2} - {\left(3 x + 1\right)}^{2} \frac{\left(\sqrt{x} + 1\right)}{\sqrt{x}}}{\sqrt{x} + 1} ^ 4$

Simplifying gives us,

$\frac{d}{\mathrm{dx}} \left({\left(3 x + 1\right)}^{2} / {\left(\sqrt{x} + 1\right)}^{2}\right) = \frac{9 {x}^{2} \sqrt{x} + 27 {x}^{2} + 18 x \sqrt{x} + 6 x + 5 \sqrt{x} - 1}{\sqrt{x} {\left(\sqrt{x} + 1\right)}^{4}}$