To find the derivative of the function \( f(x) = \sin(x)^{\frac{4}{3}} \cdot \cot(x) \), we will use the product rule for differentiation. The product rule states that if we have two functions \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is \( u'(x)v(x) + u(x)v'(x) \).
In our case, let's set \( u(x) = \sin(x)^{\frac{4}{3}} \) and \( v(x) = \cot(x) \). Now we'll need to find the derivatives \( u'(x) \) and \( v'(x) \).
1. Differentiating \( u(x) = \sin(x)^{\frac{4}{3}} \):
Using the chain rule, we derive \( u \) with respect to \( x \). The chain rule states that if we have a composite function \( h(g(x)) \), then its derivative is \( h'(g(x))g'(x) \). In this situation, we have \( h(y) = y^{\frac{4}{3}} \) and \( g(x) = \sin(x) \).
So, we have:
\( h'(y) = \frac{4}{3}y^{\frac{1}{3}} \)
and
\( g'(x) = \cos(x) \).
Thus, \( u'(x) = h'(g(x)) \cdot g'(x) = \frac{4}{3} \sin(x)^{\frac{1}{3}} \cdot \cos(x) \).
2. Differentiating \( v(x) = \cot(x) \):
The derivative of \( \cot(x) \) is \( -\csc^2(x) \). So, \( v'(x) = -\csc^2(x) \).
Now we apply the product rule:
\( f'(x) = u'(x)v(x) + u(x)v'(x) \)
\( = \left( \frac{4}{3} \sin(x)^{\frac{1}{3}} \cdot \cos(x) \right)\cot(x) + \sin(x)^{\frac{4}{3}} \left( -\csc^2(x) \right) \)
Next, we simplify the expression. Remember that \( \cot(x) = \frac{\cos(x)}{\sin(x)} \) and \( \csc(x) = \frac{1}{\sin(x)} \), we can rewrite and simplify as follows:
\( f'(x) = \frac{4}{3} \sin(x)^{\frac{1}{3}} \cdot \frac{\cos^2(x)}{\sin(x)} - \sin(x)^{\frac{4}{3}} \cdot \frac{-1}{\sin^2(x)} \)
\( = \frac{4}{3} \sin(x)^{-\frac{2}{3}} \cdot \cos^2(x) + \sin(x)^{-\frac{2}{3}} \)
We can factor out the common term \( \sin(x)^{-\frac{2}{3}} \):
\( f'(x) = \sin(x)^{-\frac{2}{3}} \left( \frac{4}{3} \cos^2(x) + 1 \right) \)
This is the derivative of the given function. Note that the result could be simplified or rewritten in several equivalent forms depending on the context or preferences for certain expressions.