Sure! Let's find the derivative of [tex]\( y = \ln (\sin (x)) \)[/tex] with respect to [tex]\( x \)[/tex].
1. Identify the function and its inner component:
[tex]\[
y = \ln (\sin (x))
\][/tex]
Here, [tex]\( \sin(x) \)[/tex] is a composite part of the natural logarithm function [tex]\( \ln \)[/tex].
2. Apply the chain rule:
The chain rule is used when differentiating a composite function. It states that if [tex]\( y = f(g(x)) \)[/tex], then:
[tex]\[
\frac{dy}{dx} = f'(g(x)) \cdot g'(x)
\][/tex]
In our case, we have [tex]\( f(u) = \ln(u) \)[/tex] and [tex]\( g(x) = \sin(x) \)[/tex].
3. Differentiate the outer function [tex]\( f(u) = \ln(u) \)[/tex]:
The derivative of [tex]\( \ln(u) \)[/tex] with respect to [tex]\( u \)[/tex] is [tex]\( \frac{1}{u} \)[/tex]:
[tex]\[
\frac{d}{du} (\ln(u)) = \frac{1}{u}
\][/tex]
4. Substitute back the inner function [tex]\( u = \sin(x) \)[/tex]:
[tex]\[
\frac{d}{dx} (\ln(\sin(x))) = \frac{1}{\sin(x)}
\][/tex]
5. Differentiate the inner function [tex]\( g(x) = \sin(x) \)[/tex]:
The derivative of [tex]\( \sin(x) \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( \cos(x) \)[/tex]:
[tex]\[
g'(x) = \frac{d}{dx} (\sin(x)) = \cos(x)
\][/tex]
6. Combine the results using the chain rule:
[tex]\[
\frac{d}{dx} (\ln(\sin(x))) = \frac{1}{\sin(x)} \cdot \cos(x)
\][/tex]
7. Simplify the expression:
[tex]\[
\frac{dy}{dx} = \frac{\cos(x)}{\sin(x)}
\][/tex]
So, the derivative of [tex]\( y = \ln (\sin (x)) \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[
\frac{dy}{dx} = \frac{\cos(x)}{\sin(x)}
\][/tex]
This final result can also be expressed as [tex]\( \cot(x) \)[/tex], but we'll leave it in the fractional form as specified.
