Derivative of $e^{x+1}$

Question

I used an online derivative calculator to check my answer to this problem, I'm not sure what I did wrong.

To me it seems like $e^{x+1} + 1 = (x+1)e^x$

The online calculator gives $e^{x+1}$ as the derivative. What is the correct answer and why?

Thanks.

Answer 1

The correct answer is:

$$e^{x+1}$$

We use the chain rule which says:

$$(f(g(x))'=f'(g(x))g'(x)$$

If we let, $f(x)=e^{x}$ and $g(x)=x+1$ then we have $f(g(x))=e^{x+1}$. Now using the chain rule and the fact that $\frac{d}{dx}e^x=e^x$ we have:

$$(e^{x+1})'=e^{x+1}(x+1)'$$

But $(x+1)'=1$ so we have:

$$(e^{x+1})'=e^{x+1}$$

Answer 2

For $$f(x) = e^{g(x)}$$

the derivative is

$$f'(x) = g'(x)e^{g(x)}$$

by the Chain Rule of differentiation, provided $g(x)$ is differentiable.

In your case, if we let $g(x) = x+1$ then $g'(x) = 1$, leaving $f(x) = e^{x+1}$ unchanged by differentiation.

Answer 3

Note that $$\frac{d}{dx}\left[e^x\right] = e^x,$$ or equivalently, $$\frac{d}{dx}\left[\exp(x)\right] = \exp(x).$$ Then by the chain rule, with the choice $f(x) = \exp(x)$ and $g(x) = x+1$, we have $$\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) g'(x) = \frac{df}{dg} \cdot \frac{dg}{dx} = \exp(x+1) \cdot \frac{d}{dx}[x+1] = \exp(x+1),$$ and for a general differentiable function $g$, we have $$\frac{d}{dx}\left[\exp(g(x))\right] = \exp(g(x)) g'(x) = g'(x) e^{g(x)}.$$

Answer 4

Let $y = e^{x+1}$ and let $u = x+1$. Then we can write, $y = e^u$. By the chain rule,

$$f'(x) = \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = \frac{d}{du}(e^u)\cdot\frac{d}{dx}(x+1) = e^u\cdot (1) = e^u = e^{x+1}.$$

Answer 5

Notice $e^{x+1}=(e)e^x$.

$$\frac{d}{dx}(e)e^x=e\frac{d}{dx}e^x=(e)e^x=e^{x+1}.$$

Answer 6

Note: $\frac {dx^n}{dx}= nx^{n-1}$. $x$ is the variable in the base, and $n$ is a constant in the exponent.

But $\frac{de^x}{dx} \ne xe^{x-1}$. $x$ is a variable the exponent, while $e$ is a constant in the base.

You derive based on "what the variable $x$ is doing". When $x$ is in the exponent it is a very different thing than when $x$ is in the base.

$\frac{de^x}{dx} = e^x$.

Taking a cute hint from lulu:

$\frac{de^{x+1}}{dx} = \frac{d(e*e^x)}{dx} = e\frac{d(e^x)}{dx} = e*e^x = e^{x+1}$.

===

Note: $\frac{d(x^n)}{dx} = \lim \frac{(x + h)^n - x^n}{h} = \lim \frac{x^n + n*x^{n-1}*h + a*x^{n-2}h^2 + ..... + h^n - x^n}{h} = \lim \frac {n*x^{n-1}*h + h^2(ax^{n-2} + ...... + h^{n-2})}{h} = \lim n*x^{n-1} + h*(ax^{n-2} +.....) = nx^{n-1}$.

An entirely different reason for $e^x$

$\frac{d(e^x)}{dx} = \lim \frac{e^{x+h} - e^x}{h} = \lim \frac{e^xe^h - e^x}{h} = \lim\frac{e^x(e^h - 1)}{h} = e^x * \lim \frac{e^h -1}{h} = e^x * 1$.

Completely different.

Answer 7

You can also use logarithmic differentiation

First set your function in terms of $x$ and $y$:

$y=e^{x+1}$

Then take the natural logarithm of both sides.

i.e

$ln(y)=ln(e^{x+1})$

$ln(y)= x+1$

Then use implicit differentiation to derive both sides of the equation

$1\over{y}$$dy\over{dx}$$=1$

Move $y$ over to the right side by multiplying both sides by $y$.

So:

$dy \over{dx}$$=y$

Which means:

$dy \over{dx}$$=e^{x+1}$

The derivative of $e^{x+1}$ is itself.

Hope this can be helpful.