Question

\frac{d}{dx}\int_{\frac{\pi}{2}}^{\sqrt{x}}\sin 13t dt = \frac{d}{dx}\left(-\frac{\cos(13\sqrt{x})}{13}\
ight)

Explanation:

Step1: Apply the fundamental theorem of calculus and chain rule

The fundamental theorem of calculus states that if \(F(x)=\int_{a}^{x}f(t)dt\), then \(F^\prime(x) = f(x)\). For \(\int_{a}^{u(x)}f(t)dt\), by the chain - rule, its derivative is \(f(u(x))\cdot u^\prime(x)\).
Here, \(a = \frac{\pi}{2}\), \(u(x)=\sqrt{x}\), and \(f(t)=\sin(13t)\). First, we know that \(\int\sin(13t)dt=-\frac{\cos(13t)}{13}+C\).

Step2: Differentiate the right - hand side

We have \(y =-\frac{\cos(13\sqrt{x})}{13}\). Using the chain rule, if \(y =-\frac{\cos(u)}{13}\) where \(u = 13\sqrt{x}\).
The derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx}=-\frac{1}{13}\cdot(-\sin(u))\cdot\frac{du}{dx}\).
Since \(u = 13\sqrt{x}\), then \(\frac{du}{dx}=\frac{13}{2\sqrt{x}}\).
Substituting \(u = 13\sqrt{x}\) into the formula, we get \(\frac{dy}{dx}=\frac{\sin(13\sqrt{x})}{13}\cdot\frac{13}{2\sqrt{x}}\).

Answer:

\(\frac{\sin(13\sqrt{x})}{2\sqrt{x}}\)