Question

If \(\mathrm{f}(\mathrm{x})=\int_{0}^{\mathrm{g}(\mathrm{x})} \frac{1}{\sqrt{1+\mathrm{t}^{3}}} \mathrm{dt}\), where \(\mathrm{g}(\mathrm{x})=\int_{0}^{\cos \mathrm{x}}\left[1+\sin \left(\mathrm{t}^{2}\right)\right] \mathrm{dt}\), find \(\mathrm{f}^{\prime}(\pi / 2)\)

Short answer

Answer: The value of the derivative of f(x) at x = π/2 is \(1 + \sin(\pi^2/16)\).

Step-by-step solution

Find the derivative of the inner integral function g(x)

In this step, we will apply the Fundamental theorem of calculus to find the derivative of g(x) with respect to x. The Fundamental theorem of calculus states that if \(G(x) = \int_0^x h(t) dt\), then \(G'(x) = h(x)\). So in our case, the derivative of g(x) with respect to x is: $$g'(x) = 1 + \sin(x^2)$$

Apply the Chain rule to find the derivative of f(x)

Now that we have the derivative of the inner integral function g(x), we can apply the Chain rule to find the derivative of f(x) with respect to x. The Chain rule states that if we have a composite function h(x) = f(g(x)), then the derivative of h(x) with respect to x is given by: $$h'(x) = f'(g(x)) \cdot g'(x)$$ Applying the Chain rule to our function, f(x), we need to find \(f'(g(x)) \cdot g'(x)\). Using the Fundamental theorem of calculus, we get that the derivative of f(x) with respect to t is $$f'(t) = \frac{1}{\sqrt{1+t^3}}$$ So, the derivative of f(x) is given by: $$f'(x) = f'(g(x)) \cdot g'(x) = \frac{1}{\sqrt{1+g(x)^3}} \cdot (1 + \sin(x^2))$$

Compute f'(π/2)

Now that we have the general formula for f'(x), we can simply substitute in the value of x = π/2 to find the derivative at that point. We know that at x = π/2, the value of g(x) is: $$g(\pi/2) = \int_{0}^{\cos \frac{\pi}{2}}\left[1+\sin \left(t^{2}\right)\right] dt$$ As the cosine of π/2 is 0, g(π/2) will be 0. Now we just need to compute f'(π/2). So, $$f'(\pi/2) = \frac{1}{\sqrt{1+g(\pi/2)^3}} \cdot (1 + \sin((\pi/2)^2))$$ $$f'(\pi/2) = \frac{1}{\sqrt{1+0}} \cdot (1 + \sin((\pi/4)^2))$$ $$f'(\pi/2) = (1)(1 + \sin(\pi^2/16))$$ $$f'(\pi/2) = 1 + \sin(\pi^2/16)$$ So, the value of the derivative of f(x) at x = π/2 is \(1 + \sin(\pi^2/16)\).