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
$\displaystyle y=mx+c$ 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
$\displaystyle y = x + 1$
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
$\displaystyle y-1 = 1(x-0)$
$\displaystyle y=x+1$