# How do you differentiate sin((2sqrtx+1)/(x+1))?

Jan 14, 2017

$y ' = \cos \left(\frac{2 \sqrt{x} + 1}{x + 1}\right) \cdot \frac{\left(x + 1\right) {x}^{- \frac{1}{2}} - \left(2 \sqrt{x} + 1\right)}{x + 1} ^ 2$

#### Explanation:

First treat $\frac{2 \sqrt{x} + 1}{x + 1}$ as a "unit" inside the sine function.

So, we will differentiate the sine function first and then apply the chain rule to differentiate the "unit" that is inside the sine function using quotient rule.

$y = \sin \left(\frac{2 \sqrt{x} + 1}{x + 1}\right)$

$y ' = \cos \left(\frac{2 \sqrt{x} + 1}{x + 1}\right) \cdot \left(\frac{2 \sqrt{x} + 1}{x + 1} \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$y ' = \cos \left(\frac{2 \sqrt{x} + 1}{x + 1}\right) \cdot \frac{\left(x + 1\right) {x}^{- \frac{1}{2}} - \left(2 \sqrt{x} + 1\right)}{x + 1} ^ 2$

Jan 14, 2017

$\cos \left(\frac{2 \sqrt{x} + 1}{x + 1}\right) \frac{x + 1 - 2 x - \sqrt{x}}{\sqrt{x} {\left(x + 1\right)}^{2}}$

#### Explanation:

Let $f \left(x\right) = \frac{2 \sqrt{x} + 1}{x + 1}$. We are trying to find (sin(f(x))'

With the chain rule, that is $\cos \left(f \left(x\right)\right) \cdot f ' \left(x\right)$

To find the derivative of $f$, use the quotient rule:

$f ' \left(x\right) = \frac{\frac{1}{\sqrt{x}} \left(x + 1\right) - 2 \sqrt{x} - 1}{x + 1} ^ 2$

The final derivative is: $\cos \left(\frac{2 \sqrt{x} + 1}{x + 1}\right) \frac{\frac{1}{\sqrt{x}} \left(x + 1\right) - 2 \sqrt{x} - 1}{x + 1} ^ 2$.

This can be simplified like this:

$\cos \left(\frac{2 \sqrt{x} + 1}{x + 1}\right) \frac{x + 1 - 2 x - \sqrt{x}}{\sqrt{x} {\left(x + 1\right)}^{2}}$