Question

If \(\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{x}^{3}} \sqrt{\cos \mathrm{t}} \mathrm{dt}\), find \(\mathrm{f}(\mathrm{x})\)

Short answer

Answer: The derivative of the function \(f(x)\) with respect to x is \(f'(x) = \sqrt{\cos(x^3)} \cdot 3x^2\).

Step-by-step solution

Identify the fundamental theorem of calculus

According to the fundamental theorem of calculus, if we have a function \(f(x) = \int_{a}^{g(x)} u(t) \mathrm{dt}\), where u(t) is a continuous function, then the derivative of f(x) with respect to x is given by \(f'(x) = u(g(x)) \cdot g'(x)\).

Differentiate f(x) with respect to x

Now, let's differentiate the given function with respect to x using the fundamental theorem of calculus. We have, \(f(x) = \int_{0}^{x^3} \sqrt{\cos t} \mathrm{dt}\). So, using the fundamental theorem of calculus, we get \(f'(x) = \sqrt{\cos(x^3)} \cdot \frac{d(x^3)}{dx}\).

Determine the derivative of x^3

In order to complete the differentiation, we need to compute the derivative of \(x^3\) with respect to x. Using the power rule, we get \(\frac{d(x^3)}{dx} = 3x^2\).

Combine the results

Now let's combine the results from step 2 and step 3. We found that \(f'(x) = \sqrt{\cos(x^3)} \cdot \frac{d(x^3)}{dx}\), and \(\frac{d(x^3)}{dx} = 3x^2\). Therefore, \(f'(x) = \sqrt{\cos(x^3)} \cdot 3x^2\).

Integrate f'(x) with respect to x to find f(x)

Finally, we need to find the original function f(x) by integrating its derivative f'(x) with respect to x: \(f(x) = \int f'(x) \mathrm{dx} = \int \sqrt{\cos(x^3)} \cdot 3x^2 \mathrm{dx}\). Unfortunately, this integral does not have a simple anti-derivative that can be expressed in terms of elementary functions. So, the original function f(x) can be described as \(f(x) = \int_{0}^{x^{3}} \sqrt{\cos t} \mathrm{dt}\), which was given in the problem.