Find the equation of the tangent to the curve Find the equation of the tangent to the curve:
$$y = e^x +1$$
At the point:
$$(1, e+1)$$
My process:
$$Gradient: y' = e^x$$
Tangent: $$y-(e+1) = e^x(x-1)$$
$$y= xe^x-e^x+e+1$$
I don't understand where my mistake is. I found the derivative (gradient) and then put it into the gradient formula to find the equation of the tangent but I am still wrong and I don't know where?

Your mistake is as follows:
$$y-y_1=m(x-x_1)$$
$$y-(e+1)=\color{red}{e^{x}}(x-1)$$
However, $m$ should be the derivative at the point where $x=1$ instead, giving:
$$y-(e+1)=e(x-1)$$
Which should give you the correct answer.

Here, I suggest an alternative approach:
Note that the equation of a tangent is given by:
$$y=mx+c \tag{1}$$
To find $m$, evaluate $y'$ of your curve $y=e^x+1$ at $x=1$.
Now you must find $c$. To do so, simply substitute $(1,e+1)$ into $(1)$ and evaluate the value of $c$.
Doing this correctly gives the following:

