AP®︎/College Calculus AB
Proving the power rule for derivatives (only the more simple cases).
The power rule tells us how to find the derivative of any expression in the form :
The AP Calculus course doesn't require knowing the proof of this rule, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.
Let's first see whether the rule makes sense for and .
Now let's prove the rule for positive integer values of .
We can also prove the rule for .
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- In the first video, he graphed the derivative of f(x). Does this mean the derivative of a function is a function itself? And does this mean you can find the derivative of a derivative?(23 votes)
- Slow your roll here, not all functions are differentiable. The function sin(1/x^2) fails to be differentiable when x = 0. If the derivative exists, then it is a function - but not all functions are differentiable! Even more frustrating is that just because a function is differentiable once does not mean it's differentiable twice. In math we denote such functions as C^1 or C^2 - where the exponent indicates how many times it's differentiable, and the each one of those resulting functions is continuous. There are even C^infinity functions such as sin, cos, and e^x.(47 votes)
- In the second video, shouldn't the first term contain n choose 0 instead of n choose 1?(5 votes)
- You are correct, however, the n choose 1 coefficient in the proof is actually for the second term in the expansion. Sal just omitted the n choose 0 coefficient for the first (x^n) term since it just multiplies the term by 1. He possibly did this to make it more obvious that it cancels with the -x^n term in the numerator.(11 votes)
- In the first video, he graphed the derivative of f(x). Does this mean the derivative of a function is a function itself? And does this mean you can find the derivative of a derivative?(4 votes)
- Yes, the derivative of a function is also a function. As long as the derivative is a differentiable function, you may indeed find the derivative of a derivative.(7 votes)
- Shouldn't the value of f'(x) be undefined at x=0 for f(x)=x^x, because 0^0 is indeterminate?(4 votes)
- Is it bad that I don't know the Binomial Theorem yet? I haven't seen it covered yet in this course.(2 votes)
- No, it's not bad. The binomial theorem is worth knowing though, because it saves time on more complicated expansions. There's a section of it on Khan Academy if you want to learn:
- where can we see the proof of ∫x dx=x2/2 + C?(0 votes)
- You can find the proof here
If you're unaware of this Integration concept till now, then you should just get a rough overview from this video as Integral calculus is a whole topic in itself(definitely related)
Hope this helps :-)(8 votes)
- What about non-positive integers. What if I want to take the derivative of a function with a negative or a fractional power?(2 votes)
- This does hold for any real number in the exponent, but showing this for general real numbers is a task more suited to an intro to real analysis course than for Calc 1.(3 votes)
- how do you differentiate a^mx(2 votes)
- Given this explanation, why is the "dx" maintained in the derivative? Such as f'(x) where f(x) = x^2 being written as 2xdx?(1 vote)
- The derivative of x^2 isn't 2xdx. It's only 2x. 2xdx would actually correspond to the differential dy (You'll learn more about differentials later, or maybe in Differential Equations).
So, if we have y = x^2, if I take the derivative w.r.t x on both sides, I get dy/dx = 2x. Here, dy/dx is the derivative, which can also be written as f'(x)(1 vote)
- Could we use newton's binomial expansion to prove it for all real numbers?(1 vote)
- I'm late, but someone else might find this helpful. You can definitely use the binomial expansion to prove it works for all real numbers, but... when you use the combination formula (n choose k), n can be a decimal or an irrational, so you'll have to take the factorial of that, but 4.5! for example, sounds impossible to find! Well, it kind of is until we introduce a new definition for factorials which works for decimals as well! (google the gamma function if you're interested)(1 vote)