help please and thank you!!

y=e^x

find the equation of the tangent line at x=0

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- Oct 30th 2008, 08:25 PMcottekrfinding the equation of a tangent line
help please and thank you!!

y=e^x

find the equation of the tangent line at x=0 - Oct 30th 2008, 08:32 PMProve It
A tangent line is a straight line, so is of the form

$\displaystyle y=mx+c$ 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.

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

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

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

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

Thus the equation of the tangent is

$\displaystyle y = x + 1$ - Oct 30th 2008, 11:44 PMWWTL@WHL
Eqn of tangent line: $\displaystyle y-y_1 = m(x-x_1)$ where $\displaystyle m$ is the gradient at the point $\displaystyle (x_1 , y_1)$

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

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

So the equation of the tangent line is

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

__$\displaystyle y=x+1$__