When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.
[Math] Why isn’t the derivative of $e^x$ equal to $xe^{(x-1)}$
calculusderivatives
Related Solutions
$$f_a(x)=f(a,x)=ax$$ $$f_a(f_a(x))=f(a,f_a(x))=f(a,f(a,x))=af(a,x)=a^2x$$ $${\partial\over\partial a}f(a,x)={\partial(ax)\over\partial a}=x$$ $${\partial\over\partial x}f(a,x)={\partial(ax)\over\partial x}=a$$ $${\partial\over\partial a}f(a,f(a,x))={\partial\over\partial a}f(a,u)$$ where,$$u=f(a,x)=ax$$ Then: $${\partial\over\partial a}f(a,u)={\partial(au)\over\partial a}+{\partial f(a,u)\over\partial u}{\partial u\over\partial a}=u+ax=ax+ax=2ax$$
I have been in your shoes at one point, and I think that your problem is that you are thinking of derivative as finding slopes. I instead I present a new abstract interpretation which will possibly dispel all doubts.
If we have an equation of the sort:
$$ x+3 = 2$$
Then we can rearrange it to get:
$$ x=-1 $$
Similarly, if have an equation with two variables (implicit curve) then we apply the $\frac{d}{dx}$ operator on both sides to find to relate the rate of change of variables. For example, consider the equation of a circle:
$$ x^2 + y^2 =1$$
If we apply $ \frac{d}{dx}$
$$ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}1$$
Now here we say that to keep on the circle, when we change our $x$ the $y$ must change as a function of it. Considering that and simplifying,
$$ \frac{d}{dx} x^2 + \frac{d}{dx} y^2 =0$$
Or,
$$ 2x + 2y y' = 0$$
Or,
$$ y' = -\frac{x}{y}$$
WIth that in mind,
$$ \frac{d}{dx} \sin y = \frac{d}{dx} ( \sin x) |_y \frac{dy}{dx}$$
Is due to the fact that we are saying that $y$ is a function of $x$. The real idea behind applying the chain rule when the $y$ is inside because we want to say that $y$ is dependent on $x$ but if we said both variables were not correlated at all i.e: you could freely change $x$ and $y$ independent of each other then the derivative would be zero.
Remembering identities
First of all I suggest that you try to derive all the identities by yourself from scratch. However, there is an easy way to derive the inverse identities by a clever application of the chain rule. I have written about it here.
For learning more about $d$ as an operator, see here
Best Answer
You are confusing things. If I define $f : \mathbb{R} \to \mathbb{R}$ by $f(x)=x^n$ this is very different from defining $g: \mathbb{R} \to \mathbb{R}$ by $g(x)= a^x$, note that in the first one the exponent is not varying and on the other function the variable appears on the exponent.
For the first function the derivative is just $f'(x) = nx^{n-1}$, for the second one things are different. First it turns out that first you need to define what it means to raise something to a real number (notice that the usual definition doesn't work, what would mean multiplying a number by itself $\pi$ times?), in that case for reasons that I won't explain here we define this function as:
$$a^x = e^{x\ln a}$$
In that case, if we know how to differentiate $e^x$ (and usually when we construct this, we already know), we'll have the following:
$$(\ln \circ g)(x)=x \ln a$$
Now the chain rule gives:
$$\ln'(g(x))g'(x)=\ln a$$
However $\ln'(x) = 1/x$ because of the construction of $\ln$ and $g(x)=a^x$ so tha we have:
$$\frac{1}{a^x}g'(x)=\ln a \Longrightarrow g'(x) = a^x \ln a$$
Notice that there was a crucial appeal to the definition $a^x = e^{x \ln a}$. To know why we define things this way look at Spivak's Calculus, there's an entire chapter devoted to all the constructions about logs and exponentials.