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The operator has to act on something (or, as I like to say, it needs something to eat!:).
What's $\sqrt{}$? (It's nothing without an argument.) What's $\sin{}$? (That's right -- sine without an argument is a sin!)
Treat limits the same way.
This proof made use of one of our new tools (the product rule). Once you build and prove a tool, it becomes a power tool:).
The power rule works for any real exponent -- it's just that we've only proven it for positive integers. You may use it for other powers (e.g. $\sqrt{x}=x^\frac{1}{2}$).
But we'll obtain this theorem by using tools we've already built; in particular, the product rule, and a trick to get the derivative of the multiplicative inverse of a function.
A theorem that you prove on the way to proving some theorem is called a "lemma".
where $e \approx 2.71828$ (it's irrational, like $\pi$).
This is certainly one of the most amazing rules. It says that $e^x$ is an exponential function which is its own derivative: whose values are its slopes, as well (and all its higher derivatives as well -- its concavity, its jerk, ...).
Here's an animation which motivates the discussion...
We'll need a couple of limits to establish this one, as well as one trig identity: \[ \sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a) \]
This is a consequence of the derivative of sine, by symmetry and periodicity.
Further consequences: \[ \frac{d^2}{dx^2}[\sin(x)]=-\sin(x) \] \[ \frac{d^3}{dx^3}[\sin(x)]=-\cos(x) \] \[ \frac{d^4}{dx^4}[\sin(x)]=\sin(x) \]
The sine (and cosine) functions are their own fourth derivatives.