# Derivative A1

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https://algebrica.org/exercises/derivative-a-1/
Fetched from algebrica.org test 6184; source modified 2025-03-07T14:42:03.

This exercise requires calculating the derivative of a [composite power function](../derivative-of-composite-power-functions.md) of the form \\( f(x)^{g(x)} \\).

Let’s consider the function \\( y = x^{2cosx} \\), and calculate its derivative.

* * *

First, let’s rewrite the function by applying the logarithm to both sides:

\\[
\\ln y = \\ln(x^{2cosx})
\\]

For the properties of logarithms \\(\\log\_a(b^c) = c \\cdot \\log\_a(b)\\)

The equality can be rewritten as:

\\[
\\ln y = 2cosx \\cdot \\ln(x)
\\]

* * *

Since \\(\\ln y\\) is a composite function, its derivative is

\\[
\\frac{1}{y} \\cdot y’
\\]

Let’s compute the derivative for the element on the right-hand side of the equality \\(2cosx \\cdot \\log(x)\\):

\\[
-2\\sin(x) \\cdot \\ln(x) + \\frac{2\\cos(x)}{x}
\\]

We obtain:

\\[
\\frac{1}{y} \\cdot y’ = -2\\sin(x) \\cdot \\ln(x) + \\frac{2\\cos(x)}{x}
\\]

* * *

The equality can be rewritten as:

\\[
y’ = y \\cdot \\left(-2\\sin(x) \\cdot \\ln(x) + \\frac{2\\cos(x)}{x} \\right)
\\]

Since \\(y = x^{2cosx}\\), we have:

\\[
y’ = x^{2cosx} \\cdot \\left(-2\\sin(x) \\cdot \\ln(x) + \\frac{2\\cos(x)}{x} \\right)
\\]

Thus, the derivative of \\( y = x^{2cosx} \\) is equal to:

\\[
y’ = x^{2cosx} \\cdot \\left(-2\\sin(x) \\cdot \\ln(x) + \\frac{2\\cos(x)}{x} \\right)
\\]