# Slope of the curve

Gold Member

## Homework Statement

Slope of the curve at the point x=0 of the function y=y(x) specified implicitly as $\displaystyle \int_0^y e^{-t^2} dt + \int_0^x \cos t^2 dt = 0$ is

## The Attempt at a Solution

Differentiating both sides wrt x
$e^{-y^2} \frac{dy}{dx} + \cos x^2 = 0$
Now If I put x=0 above
$\dfrac{dy}{dx} = \dfrac{-1}{e^{-y^2}}$

But I don't know the value of y when x = 0.

Dick
Homework Helper

## Homework Statement

Slope of the curve at the point x=0 of the function y=y(x) specified implicitly as $\displaystyle \int_0^y e^{-t^2} dt + \int_0^x \cos t^2 dt = 0$ is

## The Attempt at a Solution

Differentiating both sides wrt x
$e^{-y^2} \frac{dy}{dx} + \cos x^2 = 0$
Now If I put x=0 above
$\dfrac{dy}{dx} = \dfrac{-1}{e^{-y^2}}$

But I don't know the value of y when x = 0.

You should be able to figure it out. Put x=0 into your original integral equation.