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?

  • $\begingroup$ $y-(e+1) = e^{\color{red}{1}}(x-1)$ $\endgroup$ Mar 1, 2017 at 10:14

2 Answers 2


At $(1, e+1)$,

$$y' = e^{1} = e$$


$$y - (e + 1) = e(x - 1)$$ $$y = ex + 1$$

I think it is a careless mistake.


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:

enter image description here


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