Question:

Find the derivative of the function $f(x) = e^x \cdot \cos(x) + \log_2(x^3) - \sin^2(x)$.

Solution:

To find the derivative of a function that involves different types of basic functions, we will use the chain rule, product rule, derivative of exponential functions, derivative of logarithmic functions, and derivative of trigonometric functions.

Given function: $f(x) = e^x \cdot \cos(x) + \log_2(x^3) - \sin^2(x)$

To find the derivative of this function, we will differentiate each term separately and add them together:

Term 1: $e^x \cdot \cos(x)$

Applying the product rule, which states that for two functions $u(x)$ and $v(x)$, the derivative of their product is $u'(x) \cdot v(x) + u(x) \cdot v'(x)$.

Here, $u(x) = e^x$ and $v(x) = \cos(x)$.

We know that the derivative of $u(x) = e^x$ is simply $u'(x) = e^x$.

The derivative of $v(x) = \cos(x)$ is $v'(x) = -\sin(x)$.

Using the product rule, we have:

$\frac{d}{dx} (u(x) \cdot v(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

$\frac{d}{dx} (e^x \cdot \cos(x)) = e^x \cdot \cos(x) + e^x \cdot -\sin(x)$

Simplifying, we get:

$\frac{d}{dx} (e^x \cdot \cos(x)) = e^x \cdot \cos(x) - e^x \cdot \sin(x)$

Term 2: $\log_2(x^3)$

To find the derivative of $\log_2(x^3)$, we will use the chain rule.

Let's rewrite $\log_2(x^3)$ as $\ln(x^3) \cdot \frac{1}{\ln(2)}$ to apply the chain rule more easily.

Using the chain rule, we have:

$\frac{d}{dx} (f(g(x))) = f'(g(x)) \cdot g'(x)$

Here, $f(g(x)) = \ln(x^3)$, so $f(g(x))' = \frac{d}{dx} (\ln(x^3))$.

Let's define $g(x) = x^3$ and $f(u) = \ln(u)$.

Using the derivative of $f(u) = \ln(u)$, which is $f'(u) = \frac{1}{u}$, we have:

$\frac{d}{dx} (\ln(x^3)) = \frac{1}{x^3} \cdot \frac{d}{dx} (x^3)$

Simplifying, we get:

$\frac{d}{dx} (\ln(x^3)) = \frac{3}{x}$

Since we rewrote $\log_2(x^3)$ as $\ln(x^3) \cdot \frac{1}{\ln(2)}$, we need to multiply the derivative of $\ln(x^3)$ by $\frac{1}{\ln(2)}$:

$\frac{d}{dx} (\log_2(x^3)) = \frac{3}{x} \cdot \frac{1}{\ln(2)}$

Simplifying further:

$\frac{d}{dx} (\log_2(x^3)) = \frac{3}{x \cdot \ln(2)}$

Term 3: $-\sin^2(x)$

To differentiate $-\sin^2(x)$, we will apply the chain rule.

Let $u(x) = \sin(x)$ and $f(u) = -u^2$.

Derivative of $f(u) = -u^2$ is $f'(u) = -2u$.

Applying the chain rule, we have:

$\frac{d}{dx} (f(g(x))) = f'(g(x)) \cdot g'(x)$

Here, $f(g(x)) = \sin(x)$, so $f(g(x))' = \frac{d}{dx} (\sin(x))$.

Let's define $g(x) = x$.

Using the derivative of $\sin(x) = \cos(x)$, we have:

$\frac{d}{dx} (\sin(x)) = \cos(x)$

Multiplying by the derivative of $f(u) = -u^2$:

$\frac{d}{dx} (-\sin^2(x)) = -2\sin(x) \cdot \cos(x)$

Now, we can find the derivative of the given function $f(x)$ by summing up the derivatives of each term:

$\frac{d}{dx} (e^x \cdot \cos(x) + \log_2(x^3) - \sin^2(x)) = e^x \cdot \cos(x) - e^x \cdot \sin(x) + \frac{3}{x \cdot \ln(2)} - 2\sin(x) \cdot \cos(x)$

Thus, the derivative of $f(x)$ is $e^x \cdot \cos(x) - e^x \cdot \sin(x) + \frac{3}{x \cdot \ln(2)} - 2\sin(x) \cdot \cos(x)$.

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