# Differentiating an integral

## Homework Statement

$$\int_{2}^{y(x)}e^{t^2+1}dt + \int_{x^2+3x}^{0}\frac{e^z}{1+z}=0$$

I need so find $$y'(0)$$.

## The Attempt at a Solution

$$\frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt =y'(x) \cdot e^{y(x)^2+1}$$

$$\frac{d}{dx}\int_{x^2+3x}^{0}\frac{e^z}{1+z}=-\frac{e^{x^2+3x}}{x^2+3x+1}\cdot (2x+3)$$

adding them and substituting x=0 yields:

$$y'(0) \cdot e^{y(0)^2+1} -3=0$$

but how can i find y'(0) from them equation?

## Homework Statement

$$\int_{2}^{y(x)}e^{t^2+1}dt + \int_{x^2+3x}^{0}\frac{e^z}{1+z}=0$$

I need so find $$y'(0)$$.

## The Attempt at a Solution

$$\frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt =y'(x) \cdot e^{y(x)^2+1}$$

$$\frac{d}{dx}\int_{x^2+3x}^{0}\frac{e^z}{1+z}=-\frac{e^{x^2+3x}}{x^2+3x+1}\cdot (2x+3)$$

adding them and substituting x=0 yields:

$$y'(0) \cdot e^{y(0)^2+1} -3=0$$

but how can i find y'(0) from them equation?

$$y'(x)= \frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt + \frac{d}{dx}\int_{x^2+3x}^{0}\frac{e^z}{1+z}$$

Go with solving....

$$y'(0) = y'(0) \cdot e^{y(0)^2+1} -3$$

$$y'(0) =\frac{-3}{1-e^{y(0)^2+1}}$$

That's it.

but I need to find a number, not an expression involving y(0). how can y(0) be found?

You don't know y(x), so you can't know y'(0).