# Question #bcf6f

Mar 5, 2017

I understand the question is how the derivative of $F \left(x\right)$ is calculated, is not it? If that is so, the derivative of $F \left(x\right)$ is:

$F ' \left(x\right) = 2 \csc 2 x$.

#### Explanation:

The answer is obtained by applying the properties of the derivation and the rule of the chain. First, since our function is a subtraction between two expressions, its derivative will be the subtraction of the derivatives of both expressions separately:

$F ' \left(x\right) = \left(\ln \left(\sin x\right)\right) ' - \left(\ln \left(\cos x\right)\right) '$.

Now, we must apply the derivative of the logarithm (using the chain rule also), which is:

$y = \ln \left(f \left(x\right)\right) \Rightarrow y ' = \frac{f ' \left(x\right)}{f \left(x\right)}$,

with which we will have the following:

$F ' \left(x\right) = \frac{\left(\sin x\right) '}{\sin x} - \frac{\left(\cos x\right) '}{\cos x}$.

Finally, we derive the sine and cosine:

$F ' \left(x\right) = \frac{\cos x}{\sin x} - \frac{- \sin x}{\cos x} = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}$.

The derivative is already calculated. We can leave the result or try to simplify it a little ...

We add the two fractions and use the fundamental theorem of trigonometry to simplify the fraction obtained:

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

We can improve it a little more, remembering the formula of the sine of the double angle:

$\sin 2 \alpha = 2 \sin \alpha \cdot \cos \alpha$.

Using this formula, the expression of the derivative may be simplified as follows:

$F ' \left(x\right) = \frac{1}{\sin x \cdot \cos x} = \frac{1}{\frac{\sin 2 x}{2}} = \frac{2}{\sin 2 x} = 2 \csc 2 x$.