# Thread: finding the equation of a tangent line

1. ## finding the equation of a tangent line

y=e^x

find the equation of the tangent line at x=0

2. Originally Posted by cottekr

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

$y = x + 1$

3. 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

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

$y=x+1$