Derivative A1

Source: algebrica.org - CC BY-NC 4.0 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 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)