# Question 217eb

Nov 30, 2016

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

#### Explanation:

First of all, the function can be simplified.

Call the function $f \left(x\right)$.

$f \left(x\right) = \frac{x \tan x}{\frac{1}{\cos} x + \sin \frac{x}{\cos} x}$

$f \left(x\right) = \frac{x \tan x}{\frac{1 + \sin x}{\cos} x}$

$f \left(x\right) = \frac{x \tan x \left(\cos x\right)}{1 + \sin x}$

$f \left(x\right) = \frac{x \sin \frac{x}{\cos} x \left(\cos x\right)}{1 + \sin x}$

$f \left(x\right) = \frac{x \sin x}{1 + \sin x}$

We now differentiate this using the quotient rule. The quotient rule states that for a function color(red)(f(x) = (g(x))/(h(x)), the derivative is given by color(red)(f'(x) = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2#. We know the derivative of the numerator is $\sin x + x \cos x$ (by the product rule), and that the derivative of the denominator is $\cos x$.

Hence:

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

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

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

Hopefully this helps!