Question

If \[ y(x) = \int_{\sqrt{x}}^x e^t \, dt, \quad x>0 \] then \( y'(1) = \) .............

Hint:

When differentiating integrals with variable limits, use the Leibniz rule to handle both the upper and lower limits.

Answer 1

Correct Answer: 1.34 - 1.36

Step 1: Differentiating using the Fundamental Theorem of Calculus.
We differentiate \( y(x) \) using the Leibniz rule for differentiating an integral with variable limits: \[ y'(x) = e^x - e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}. \]
Step 2: Evaluating at \( x = 1 \).
Substituting \( x = 1 \) into the derivative: \[ y'(1) = e^1 - e^{\sqrt{1}} \cdot \frac{1}{2\sqrt{1}} = e - \frac{e}{2} = 0. \]