What is the partial derivative of e^(x-1) with respect to x.

Question

I really got confused about it. And need some help

Answer 1

As mentioned, partial derivatives are for functions of more than one variable. So we write,

$$\frac{d}{dx} e^{x-1}$$

Doesn't need chain rule, though:

$$\frac{d}{dx} e^{x-1} = \frac{d}{dx} \frac{e^{x}}{e} = \frac{1}{e} \frac{d}{dx} e^{x} = \frac{e^{x}}{e}$$

But if you had,

$$\frac{d}{dx} e^{x-1} = e^{x-1} \cdot \frac{d}{dx} (x-1) = e^{x-1} \cdot 1$$

Answer 2

Let $\mathbf{x}=(x_1,...,x_n)$ be a vector, then the definition of partial derivative is
$$\frac{\partial f(\mathbf{x})}{\partial x_i}=\lim_{\xi\to0}\frac{f(\mathbf{x}+\xi \mathbf{e}_i)-f(\mathbf{x})}{\xi}$$ where $\mathbf{e}_i$ is the standard basis. The computation is similar to a one variable derivative,in which every other variable that is not being differentiated is held constant. Suppose you have a surface described by $f(\mathbf{x})$, then for any point on this surface, there are an infinite number of tangent lines. Loosely speaking, when you take a partial derivative you are effectively finding the slope of one of these lines. Other frequently used notations are $f_{x_i},\partial_{x_i}f$ and $D_{x_i}f$. As mentioned, this is an extension of functions defined on $\mathbb R$.

Now considering your case, $$f(\mathbf{x})=e^{x_1-1}$$ but $\mathbf{x}=(x_1)$ and so it would be more convenient to use one variable differentiation in $\mathbb R$, since partial differentiation is an extensionin $\mathbb{R}^n$, then $$\lim_{\xi\to0}\frac{f(x+\xi)-f(x)}{\xi}=\frac{df(x)}{dx}.$$ As other have mentioned, using the Chain Rule, let $u=x-1$, $$\frac{df(x)}{dx}=\frac{df(u)}{dx}=\frac{e^u}{du}\frac{du}{dx}=e^u\frac{d(x-1)}{dx}=e^u=e^{x-1}.$$

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You can also apply the definition

$$\frac{df}{dx}=\lim_{\xi\to0}\frac{e^{x+\xi-1}-e^{x-1}}{\xi}=e^{x-1}\lim_{\xi\to0}\frac{e^{\xi}-1}{\xi}=e^{x-1}$$