如何求积分的导数？求解特定变上限积分的导数计算难题

Hey there! Great call recognizing that integrating first is a dead end—this integral ties to the error function, which doesn’t have an elementary closed-form antiderivative. Instead, we can use a combination of the Fundamental Theorem of Calculus (Part 1) and the chain rule to compute this derivative directly, no integration required. Let’s walk through it step by step:

Step 1: Factor out the constant #

First, pull the constant coefficient outside the derivative (differentiation is linear, so this is totally valid):
$$\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac 2{\sqrt{2\pi}}\int_0^{\sqrt y} \exp\left(-{\frac{x^2}{2}}\right) ,\mathrm{d}x\right) = \frac{2}{\sqrt{2\pi}} \cdot \frac{\mathrm{d}}{\mathrm{d}y}\left(\int_0^{\sqrt y} \exp\left(-\frac{x^2}{2}\right) dx\right)$$

Step 2: Apply chain rule with FTC Part 1 #

Let’s set ( u = \sqrt{y} = y^{1/2} ). Our integral becomes ( \int_0^u \exp\left(-\frac{x^2}{2}\right) dx ). By the Fundamental Theorem of Calculus Part 1, the derivative of this integral with respect to ( u ) is just the integrand evaluated at ( u ):
$$\frac{\mathrm{d}}{\mathrm{d}u}\left(\int_0^u \exp\left(-\frac{x^2}{2}\right) dx\right) = \exp\left(-\frac{u^2}{2}\right)$$

Since we need the derivative with respect to ( y ), use the chain rule: multiply the above result by ( \frac{du}{dy} ):
$$\frac{\mathrm{d}}{\mathrm{d}y}\left(\int_0^{\sqrt y} \exp\left(-\frac{x^2}{2}\right) dx\right) = \exp\left(-\frac{(\sqrt{y})^2}{2}\right) \cdot \frac{\mathrm{d}}{\mathrm{d}y}(\sqrt{y})$$

Step 3: Simplify the individual terms #

Break down the two parts:

• ( (\sqrt{y})^2 = y ), so the exponential term simplifies to ( \exp\left(-\frac{y}{2}\right) )
• The derivative of ( \sqrt{y} ) is ( \frac{1}{2\sqrt{y}} )

Step 4: Combine and clean up the result #

Multiply all the pieces together and cancel the 2s:
$$\frac{2}{\sqrt{2\pi}} \cdot \exp\left(-\frac{y}{2}\right) \cdot \frac{1}{2\sqrt{y}} = \frac{\exp\left(-\frac{y}{2}\right)}{\sqrt{2\pi y}}$$

That’s your final simplified derivative! This approach skips the impossible task of computing the integral first, which is the standard trick for derivatives of variable-limit integrals.

内容的提问来源于stack exchange，提问作者Michael