finding the equation of a tangent line

**cottekr** · October 30th 2008, 08:25 PM

help please and thank you!!

y=e^x

find the equation of the tangent line at x=0

**Prove It** · October 30th 2008, 08:32 PM

Originally Posted by cottekr

help please and thank you!!

y=e^x

find the equation of the tangent line at x=0

A tangent line is a straight line, so is of the form

$y=mx+c$ where m is the gradient and c is the y-intercept.

To find the gradient of this tangent, we need to differentiate the function and evaluate it at that point.

$y=e^x$ so $\frac{dy}{dx} = e^x$ .

At the point $x=0$ , $\frac{dy}{dx} = e^0 = 1$ .

So the gradient of the tangent at $x=0$ is 1.

We know that when $x=0, y=1$ , so the y-intercept is 1.

Thus the equation of the tangent is

$y = x + 1$

**WWTL@WHL** · October 30th 2008, 11:44 PM

Eqn of tangent line: $y-y_1 = m(x-x_1)$ where $m$ is the gradient at the point $(x_1 , y_1)$

When $x_1=0$ , y= $e^0 = 1$

$\frac{dy}{dx} = e^x$ at $x=0, \frac{dy}{dx} = 1$

So the equation of the tangent line is

$y-1 = 1(x-0)$

$y=x+1$

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