Question

Find \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) if \(\mathrm{y}=\int_{\mathrm{x}}^{13} \frac{\mathrm{t}^{3} \sin 2 \mathrm{t}}{\sqrt{1+3 \mathrm{t}}} \mathrm{dt}\)

Short answer

Answer: \(\frac{d^2y}{dx^2} = \frac{3x^2\sin{2x}(1+3x) + 2x^3\cos{2x}(1+3x) - \frac{3}{2}x^3\sin{2x}}{1+3x}\)

Step-by-step solution

Find the first derivative dy/dx using the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus tells us that if \(y = \int_{g(x)}^{h(x)} f(t) dt\), then \(\frac{dy}{dx} = f(h(x))h'(x) - f(g(x))g'(x)\). In this case, \(g(x) = x\), \(h(x) = 13\), and \(f(t) = \frac{t^3\sin{2t}}{\sqrt{1+3t}}\). The derivatives of \(g(x)\) and \(h(x)\) are \(g'(x) = 1\) and \(h'(x) = 0\), respectively. Applying the theorem: \(\frac{dy}{dx} = f(13)(0) - f(x)(1)\) \(\frac{dy}{dx} = -\frac{x^3\sin{2x}}{\sqrt{1+3x}}\)

Find the second derivative d^2y/dx^2 by taking the derivative of the first derivative

Now we must differentiate \(\frac{dy}{dx}\) with respect to \(x\). Using the quotient rule, \(\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{vu'-uv'}{v^2}\), where \(u = x^3\sin{2x}\) and \(v = \sqrt{1+3x}\). Firstly, we calculate \(u'\) and \(v'\): \(u' = \frac{d}{dx}(x^3\sin{2x}) = 3x^2\sin{2x} + 2x^3\cos{2x}\) \(v' = \frac{d}{dx}(\sqrt{1+3x}) = \frac{3}{2\sqrt{1+3x}}\) Applying the quotient rule: \(\frac{d^2y}{dx^2} = \frac{\sqrt{1+3x}(3x^2\sin{2x} + 2x^3\cos{2x}) - x^3\sin{2x}(\frac{3}{2\sqrt{1+3x}})}{(1+3x)}\) \(\frac{d^2y}{dx^2} = \frac{3x^2\sin{2x}(1+3x) + 2x^3\cos{2x}(1+3x) - \frac{3}{2}x^3\sin{2x}}{1+3x}\) Therefore, the second derivative of y with respect to x is: \(\frac{d^2y}{dx^2} = \frac{3x^2\sin{2x}(1+3x) + 2x^3\cos{2x}(1+3x) - \frac{3}{2}x^3\sin{2x}}{1+3x}\)