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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 2

Lesson 7: Derivative rules: constant, sum, difference, and constant multiple: connecting with the power rule# Differentiating integer powers (mixed positive and negative)

Let's explore how to differentiate the function g(x) = 2/(x³) - 1/(x²) and evaluate the derivative at x = 2. By applying the Power Rule and simplifying, we efficiently find the slope of the tangent line at the given point.

## Want to join the conversation?

- I don't understand why if we calculate the actual expression then differentiate it the result will be wrong! I mean if we do this: 2x^2 - x^3 / x^5 . then If we differentiate this the result is like 4x - 3x^2 / 5x^4 , if we input 2 here the final result will be -1/20! Why it is like this?(7 votes)
- I am not too sure what you are asking.

But I*can*tell you that:

d/dx[2x² - (x³ / x⁵)] = // given

d/dx[2x²] - d/dx[(x³ / x⁵)] = // via derivative of sum equals sum of derivative

d/dx[2x²] - d/dx[(x³ xˉ⁵)] = // via definition of negative exponent

d/dx[2x²] - d/dx[(xˉ²)] = // via exponent multiplication "power rule"

4x - (-2xˉ³) = // take the derivative

4x + 2/x³ // via definition of negative exponent

What you appear to have done with d/dx[(x³ / x⁵)] is taken the derivative of the numerator and denominator independent of each other: (x³ / x⁵) --> 3x² / 5x⁴.

This is an illegal move, the derivative of a quotient*is not*equal to the quotient of the derivatives.

You would either need to use the quotient rule:

https://www.khanacademy.org/math/calculus-home/taking-derivatives-calc/quotient-rule-calc/v/quotient-rule-from-product-rule

or the product rule:

https://www.khanacademy.org/math/calculus-home/taking-derivatives-calc/product-rule-calc/v/applying-the-product-rule-for-derivatives

It iseasier to simplify x³ / x⁵ into xˉ²**way**

Let me know if you need more info.(18 votes)

- he has never proved the validity of the power rule for negative powers!(0 votes)
- Do you doubt its validity?

It's not difficult to prove. Start with the differential of x⁻¹ (=1/x)

d/dx (x⁻¹) = Lim (h -> 0) (1/(x+h) - 1/x)/h

= (x - (h+x))/((x+h)xh)

= -h/((x+h)xh) = -1/((x+h)x)

Which as h -> 0 = -1/x²

This agrees with the power rule with n = -1

Now once we have that result we can combine it with the chain rule and the power rule for all other negative powers. eg

x⁻ⁿ = (xⁿ)⁻¹

So d/dx (x⁻ⁿ) = -1 · n·xⁿ⁻¹/(xⁿ)² = -n/x²ⁿ⁻⁽ⁿ⁻¹⁾

= -n/xⁿ⁺¹

Which again agrees with the power rule.(15 votes)

- why does sal leace negative exponents in videos, for example at3:52in this video?(3 votes)
- Well, Sal calculates the negative exponents in the next step. Additionally, he writes the negative exponents out because it can be challenging to calculate them in your head for some.(3 votes)

- in the first step, can I write dg/dx instead of d/dx(g(x)) ?(2 votes)
- Well, yeah, they're pretty much the same; Sal wrote (d/dx)(g(x)) for the sake of his explanation.

If anyone's confused about it, just think that: (d/dx)(f(x)) means: "Take the derivative of the f(x)", while df/dx means: "The derivative of f with respect to x", you could say that the latter is a more "simplified" version of the former.(3 votes)

- Can you tell me what is the derivative of square root of x+1 / x-1 with respect to x is?(2 votes)

## Video transcript

- [Voiceover] So we have
the function g of x, which is equal to 2/x to
the third minus 1/x squared. And what I wanna do in this video, is I wanna find what g prime of x is and then I also wanna
evaluate that at x equal two, so I wanna figure that out. And I also wanna figure out what does that evaluate to when x is equal to two? So what is the slope
the of the tangent line to the graph of g, when x is equal to two? And like always, pause this video and see if you can work this out on your own before I work through it with you. And I'll give you some hints all you really need to do is apply the power rule, a little bit of basic exponent properties and some basic derivative properties to be able to do this. Alright, now let's just do this together and I'll just rewrite it. G of x is equal to this first term here, 2/x to the third. Well, that could be rewritten as 2 times x to the negative three. We know that 1/x to
the n is the same thing as x to the negative n. So I just rewrote it and this might be ringing a bell of how the power rule might be useful. And then we have minus- well, 1/x squared that is the same thing as x to the negative two. And so this, if we're gonna take the derivative of both sides of this, let's do that. Derivative with respect to x. Dx, we're gonna do that
on the left-hand side, we're also gonna do it
on the right-hand side. On the left-hand side, the derivative with respect to x of g of x, we can write that as g prime of x is going to be equal to, well, the derivative of this first that we have right here written in green, this is going to be, we're just gonna apply the power rule. We're going to take our exponent, multiply it by our coefficient out front, actually let me write that out, that's going to be... There's this equal sign. That is going to be two
times negative three, times x and now we're going
to decrement this exponent. You have to be very careful here because sometimes your brain might say, "Okay, one less than three is two, "so maybe this is x in the negative two." but remember, you're going down. So if you're at negative
three and you subtract one, we're gonna go with the negative three minus one power. We'll that's gonna take
us to negative four. So this is x to the negative four power. So two times negative three
x to the negative four, or we could have also written that as negative six x to the negative four power. And then, minus... Well, we're gonna do the same thing again right over here. We take this negative two, multiply it by the coefficient that's implicitly here, you could say there's a one there. So negative two times one. So you have the negative two there and then you have the x to the- well what's a negative two minus one? That's negative three. To the negative three power. And so we can rewrite
all of this business as, the derivative g prime of x, is equal to negative six, negative six x to the negative fourth. And now we're subtracting a negative. So we could just write this as, plus two x to the negative three. This negative cancels
out with that negative. Subtract a negative, same
thing as adding a positive. So we did the first part. We can express g prime
of x as a function of x. Now, let's just evaluate
what g prime of two is. So g prime of two is going to be equal to negative six times two to
the negative fourth power plus two times two to
the negative third power. Well, what's this going to be? This is equal to negative
6/2 to the fourth, plus 2/2 to the third, which is equal to negative six over- two to the fourth is 16, plus 2/2 to the third is eight. And so let's see, this is... Lets rewrite this all
with a common denominator. I could write this as 1/4, but then this one won't
work out as cleanly, I could write them both as eights. This is negative 3/8s. Negative 3/8s. So you have negative 3/8s plus two eights is equal to negative 1/8. So the slope of the tangent
line at x equals two, to the graph y equals g of x has a slope. That slope is negative 1/8.