Question

In Exercises \(1-20,\) find \(d y / d x\). $$y=\int_{2}^{5 x} \frac{\sqrt{1+u^{2}}}{u} d u$$

Short answer

\(dy/dx\) = \(\sqrt{1/x^{2}+25}\)

Step-by-step solution

Identify part to be differentiated

The function \(y\) to be differentiated is defined as an integral. Only the upper limit includes \(x\), that is \(5x\). So that's the part we will differentiate.

Apply Fundamental Theorem of Calculus Part 1

According to the Fundamental Theorem of Calculus Part 1, derivative of the integral of a function from a to f(x) is equal to f(x) times derivative of f(x) w.r.t \(x\) inside function: \(d/dx \int_{a}^{f(x)} g(u) du = g(f(x)) \cdot f'(x)\), Here, \(g(u) = \frac{\sqrt{1+u^{2}}}{u}\) and \(f(x) = 5x\).

Apply the theorem

By applying the theorem, we get: \(dy/dx = \frac{\sqrt{1+(5x)^{2}}}{5x} * 5 = \frac{\sqrt{1+25x^{2}}}{x}\)

Simplify the result

Simplify fraction by cancelling out the terms to get: \(dy/dx = \sqrt{1/x^{2}+25}\)