Single Variable Calculus, Chapter 5, 5.3, Section 5.3, Problem 50

Determine the derivative of the function $\displaystyle y = \int^{5x}_{\cos x} \cos (u^2) du$

Apply Properties of Integral


$
\begin{equation}
\begin{aligned}

& \int^c_a f(x) dx + \int^b_c f(x) dx = \int^b_a f(x) dx, \text{ So we have}
\\
\\
& y = \int^0_{\cos x} \cos (u^2) du + \int^{5x}_0 \cos (u^2) du
\\
\\
& y = - \int^{\cos x}_0 \cos (u^2) du + \int^{5x}_0 \cos (u^2) du

\end{aligned}
\end{equation}
$


Let $\displaystyle u_1 = \cos x_1 \frac{du_1}{dx} = - \sin x$ and $\displaystyle u_2 = 5x, \frac{du_2}{dx} = 5$, then


$
\begin{equation}
\begin{aligned}

y' =& - \frac{d}{dx} (\int^{\cos x}_0 \cos (u^2) du + \int^{5x}_0 \cos (u^2) du) \frac{du}{dx}
\\
\\
y' =& - \cos (u^2_1) \frac{du_1}{dx} + \cos (u^2_2) \frac{du_2}{dx}
\\
\\
y' =& (- \cos (\cos x)^2 \cdot - \sin x) + (\cos (5x)^2 \cdot 5)
\\
\\
y' =& \cos (\cos^2 x) \sin x + 5 \cos (25x^2)

\end{aligned}
\end{equation}
$