Question #bcf6f

1 Answer
Mar 5, 2017

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

#F' (x) = 2 csc 2x#.


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' (x) = (ln (sin x))' - (ln (cos x))'#.

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

#y = ln (f (x)) rArr y '= {f' (x)}/{f (x)}#,

with which we will have the following:

#F' (x) = {(sin x)'}/{sin x} - {(cos x)'}/{cos x}#.

Finally, we derive the sine and cosine:

#F' (x) = {cos x}/{sin x} - {- sin x}/{cos x} = {cos x}/{sin x} + {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' (x) = {cos x}/{sin x} + {sin x}/{cos x} = {cos^2 x + sin^2 x}/{sin x cdot cos x} = 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' (x) = 1/{sin x cdot cos x} = 1/{{sin 2 x}/2} = 2/{sin 2x} = 2 csc 2x#.