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### Course: AP®︎/College Calculus AB > Unit 3

Lesson 6: Selecting procedures for calculating derivatives: strategy# Manipulating functions before differentiation

Sometimes, before differentiating a function, we can rewrite it so the process of differentiation is faster and easier.

## Want to join the conversation?

- At3:00, isn't the equation not differentiable at x = 1? Do you need to make a note of "for x =/= 1" when defining the derivative in this way?(48 votes)
- Good spot! You are correct, Sal should have specified that the result is true for
`x ≠ 1`

.

I've submitted a "clarification".(39 votes)

- At7:09, x^-2*x^1/2 is x^-1/2 not x^-3/2. Not a major mistake, but just thought I'd let you know(0 votes)
- The video is correct.

x^(-2) * x^(1/2)

= x^(-2 + 1/2)

= x^(-4/2 + 1/2)

= x^(-3/2)(16 votes)

- in the second last example, why couldn't you simply use the power rule, instead of the power rule AND the chain rule? Thanks :)(2 votes)
- When the function consists of a function inside a function that cannot be simplified algebraically, then the chain rule must be used.(5 votes)

- I've got a slightly unrelated question here, how does one factorize that quickly? When I see
`x^2 + x - 2`

I solve it like this:`x^2 - x + 2x - 2`

x(x - 1) + 2(x - 1)

(x + 2)(x - 1)

I have no clue how one instantly does it...I guess knowing that I convert x into 2x - x I should probably get 2 and -1 but I think there will be cases for which it won't be true? So what's the consistent way of doing it instantly while being sure that it works?(2 votes)- When the squared term has no coesfficient, it's quite easy: find two numbers that multiplied give you the constant and that when added give you the coefficient of x. Then arrange them into (x+a)(x+b). Be sure to keep the correct coefficients. In yours, the numbers that multiplied are -2 are 1, -2 and -1, 2. The sums of these are -1 and 1, respectively. The answer easily found to be -1 and 2, for the same answer as you got. Hope that helps!(4 votes)

- are there any case where the qoutient rule is better or is it completely useless?(1 vote)
- You don't need to use the quotient rule, but it is usually easier than using a combination of the product, chain, and power rules.(4 votes)

- For the last example (at9:55) why doesn't the power rule work? When I tried it on my own, I brought the 2 to the front and changed the exponent to 1, making it 2(2x+1)^1. This gives 4x+2, which is the wrong answer.(2 votes)
- Here the expression (2x+1)^2 comprises of a function inside another function. That is: f(x)= 2x+1 and g(x)= x^2, so g(f(x))= (2x+1)^2.

So, here the chain rule is applied by first differentiating the outside function g(x) using the**power rule**which equals 2(2x+1)^1, which is also what you have done. This is then multipled by the derivative of the inside function f(x) that is 2x+1 which is 2.

So, the derivative of (2x+1)^2 = 2(2x+1)*2 = 4(2x+1) = 8x+4 as shown in10:28(1 vote)

- I got stuck at the first point if anyone can help me I will be thankful, in the
**y=x^2-x-2/x-1**, and after we simplified it gives us**d/dx[X+2]* it means if we take the derivative of it it will give us [1 ] it means for any x equal any number it doesn't matter because the result will be ONE for example if we take *x=6**for**y=x^2-x-2/x-1**the slope of the tangent line will be ONE. if x=2,3,4,5,6 the slope of the tangent line will be ONE according to instructor cuz**d/dx[X+2] =1**.

and using the quotient rule I got *dy/dx=[x^2-x-2/x-1] = X^2-2X+3/(X-1)^2 * this seems more reasonable to me cuz I can now substitute X with any value and it will give me the slope of the tangent line at that point so why that difference

I know limits using limit after simplified it will be x+2 for X not equal to 1 cuz if the x is equal to one in the denominator it will give us vertical asymptote hence there is no limit at this point thus there no derivative at that point(1 vote)- Your question is not well written. You need to use brackets remember BODMAS.

Your calculation using quotient rule seems to be incorrect. As a double check use differentiation calculator.

By the way there seems to be typo in your question. Remember how to factor quadratics.(1 vote)

- It seems as though the 2nd example at4:00is incorrect.

I keep getting x^2+5/x^2

Am I missing something or did Sal slip up?(1 vote)**You are correct**, but you miss the last step.

Quotient rule =(f'(x)g(x)-f(x)g'(x))/g(x)^2

Lets define every term

f(x)=x^2+2x-5

f'(x)=2x+2

g(x)=x

g'(x)=1

Now plug everything back in the formula

=(2x+2)*x-(x^2+2x-5)*1)/x^2

=(2x^2+2x-x^2-2x+5)/x^2

=(x^2+5)/x^2

Which equals 1+5x^(-2) -- like the video at5:37

Sal's way is preferred in my opinion(1 vote)

- Can I do a long division before differentiation to simplify functions? For example, in case if I have...

(x^3-4x^2+2x+8)/(x2-3x)

something like that...(0 votes)- It really depends on the function. Long division can potentially change a function. For example, it can't be done here. Let me explain with a simpler example.

Let's say we have the function f(x) = (x^2 - 4x)/(x-4). Now, this can be simplified by long division by canceling x-4. However, this is a new function g(x) = x. The reason I say it is a new function is that f(4) does not exist, but g(4) does exist. This makes them different functions, with potentially different derivatives.

Of course, this is not always the case. If the denominator is a function like h(x)= x^2 + x + 1 which has no real factors, then yes, cancellation is allowed because, h(x) will never equal zero, (in fact, h(x) is greater than 3/4 for all real x) and so, if it is canceled out (by a similar term in the numerator), then the function's integrity is maintained.

To sum up, long division can be applied if the denominator does not equal zero for any real number. Therefore, it is inapplicable here.(2 votes)

- For the last example you could just write (2x +1)(2x + 1) then have the realization that the applying product rule would simplify the expression to 2(f'(x)(f(x)). Ahhh the beauty of mathematics.(0 votes)

## Video transcript

- [Instructor] What I have
listed here is several of the derivative rules that
we've used in previous videos. If these things look unfamiliar
to you I encourage you to maybe to not watch this
video because in this video we're going to think about
when do we apply these rules? What strategies and can
we algebraically convert expressions so that we
can use a simpler rule? But this is a quick review,
this is of course the power rule right over here, very
handy for taking derivatives of X raised to some power. It's also we can use that
with the derivative properties of sums of derivatives or
differences of derivatives to take derivatives of polynomials. This right over here is the product rule. If I have an expression that
I want to take the derivative of and I can think of it as
the product of two functions, well then the derivative is
going to be the derivative of the first function times
the second function plus the first function times the
derivative of the second. Once again if this looks
completely unfamiliar to you or you're a little shaky,
go watch the videos, do the practice on the power
rule and the product rule, or in this case the quotient rule. And the quotient rule is
a little bit more involved and we have practicing videos
on that and I always have mixed feelings about it
because if you don't remember the quotient rule, you can
usually or you can always convert a quotient into a product
by expressing this thing at the bottom as F of X times
G of X to the negative one. So you could take the
derivative with a combination of the products and this
fourth rule over here, the chain rule. And if any of this is looking
unfamiliar again don't watch this video, this video is
for folks who are familiar with each of these derivative
rules or derivative techniques and now want to
think about well what are strategies for deciding
when to apply which. So let's do that. Let's say that I have the
expression, let's say I'm interested in taking the
derivative of X squared plus X minus two over X minus one. Which of these rules or
techniques would you use? Well you might immediately
say hey look this looks like a rational expression, I
could say this is my F of X right over here, I could
say this is my G of X right over here and I could
apply the quotient rule, this looks like a quotient
of two expressions. And you could do that and if
you do all the mathematics correctly, you will
get the correct answer. But in this case it's good
to just take a little time to realize well can I
simplify this algebraically so maybe I can do a little bit less work? And if you look at it that
way, you might realize wait what if I factored this
numerator I can factor it as X plus two times X minus one. And then I could cancel these
two characters out and I can say hey you know what this
is going to be the same thing as the derivative with
respect to X of X plus two. Derivative with respect to X
of X plus two, which is much much much much more straightforward
than trying to apply the quotient rule. Here you would just take
the derivative with respect to X of X which is just going
to be one and the derivative with respect to X of two
is just going to be zero and so all of this is
going to simplify to one. If we're taking the derivative
of that, you're essentially just using the power rule. And so once again just a
simple algebraic recognition things become much more simple. Let's do another example. So let's say that you were to
see, or someone were to ask you to take the derivative
with respect to X of, let me see, so let's say
you had X squared plus two X minus five over X. So once again you might be
tempted to use the quotient rule, this looks like the
quotient of two expressions. But then you might realize
there's some algebraic manipulations I could
do to make this simpler. You could express this as a
product, you could say that this is the same thing as,
and I'm just going to focus on what's inside the parentheses
or inside the brackets, this is the same thing as X
to the negative one times X squared plus 2x minus five
and then you might want to apply the product rule. But there's even a better
simplification here. You could just divide each of
these terms by X or one way to think about it distribute
this one over X across all the terms, X to the
negative one is the same thing as one over X and if you do
that X squared divided by X is going to be X. 2x divided by X is going to
be two and then negative five divided by X, well you could
write that as negative five over X or negative five
X to the negative one. And now I'm taking the
derivative of this with respect to X is much easier than
using either the quotient or the power rule. This is going to be, let's
see the derivative of that is going to be one, derivative
of two is going to be zero and here even though you
have a negative exponent, it might look a little
intimidating, this is just taken using the power rule. So negative one times
negative five is positive five X to the, if we take one
less than negative one we're going to go the negative two power. So once again making this
algebraic recognition simplified things a good bit. Let's do a few more examples
of just starting to recognize when we might be able to
simplify things to do things a little bit easier. So let's say that someone said
hey you take the derivative with respect to X and I'm using
X as our variable that we're taking the derivative of
with respect to but obviously this works for any
variables that we are using. So let's say we're saying
square root of X over X squared. Pause this video and think
about how would you approach this if you want to take the
derivative with respect to X of the square root of X over X squared. Well once again you might
say this is a quotient of two expressions, might try to
apply the quotient rule, or you might recognize well
look this is the same thing, let me just focus on what's
inside the brackets, you could view this as X to the negative
two times the square root of X and then you might
want to use the product rule but you could simplify this even better. You could say this is the same
thing as X to the negative two times X to the one half
power and now just using our exponent properties, negative
two plus one half is negative three halves, so this is the
derivative X to the negative three halves power. And so here once again we
took something that we thought we might have to use the
quotient rule or use the product rule and now this just becomes
a straightforward using the power rule. So this is just going to be
equal to, so bring the negative three halves out front,
negative three halves, X to the negative three halves
minus one is negative five halves power. So once again just before you,
especially if you're about to apply the quotient rule
and sometimes even the product rule, just see is there an
algebraic simplification, sometimes a trigonometric
simplification that you can make that eases your job that
makes things less hairy? As a general tip I can't say
this is going to be always true but if you're taking some
type of exam and you're going down some really hairy route
which the quotient rule will often take you, it's a
good sign that hey take a pause before trying to run
through all of that algebra to apply the quotient
rule and see if you can simplify things. So let's give another example. In this one there's not an
obvious way and it really depends on what folks'
preferences are, but let's say you want to take the derivative
with respect to X of one over 2x to the negative five,
sorry, one over 2x minus five I should say. Well here you could immediately
apply the quotient rule here the numerator you view that as F of X. You could view this as the
same thing as the derivative with respect to X. Instead of 2x minus five, let
me do that in the blue color. 2x minus five to the negative one power. In this situation, you would
use a combination of the power rule and the chain rule. You would say okay my G of X
is 2x minus five and F of G of X is going to be this whole expression. And so if you applied the
chain rule, this is going to be the derivative of the
outside function, our F of X with respect to the inside function. The derivative of F of G of
X with respect to G of X. So it's going to be negative,
we'll bring that negative out front, so we're
essentially just going to use the power rule here. Negative 2x minus five to
the negative two and then we multiply that times the
derivative of the inside function. So the inside function's
derivative, the derivative of 2x is two, the derivative
of negative five is zero so it's going to be times two
and of course you can simplify so it's a negative two
times all of this business. Let me do one more example
here just to hit the point home and once again there isn't a
must way, there isn't a way that you have to do this,
but just let you appreciate that there's multiple ways
to approach these types of derivatives. So let's say someone said
take the derivative of 2x plus one squared. Pause the video and think
about how you would do that. Well one way to do it is
just to apply the chain rule just like we just did. So you could say alright here's
going to be the derivative of the outside with respect to the inside. So it's going to be two times
2x plus one to the first power, taking one less than
that times the derivative of the inside which is just
going to be two and so this is going to be equal to
four times 2x plus one, which is equal to, if we
want to distribute the four, we could say it's 8x plus four. That's a completely
legitimate way of doing it. Now there are other ways of doing it. You could expand out
to X plus one squared. You could say hey this is the
same thing as the derivative with respect to X of 2x squared
is going to be 4x squared and then two times the product
of these terms is going to be plus 4x plus one. And now you would just
apply the power rule. It's a little bit of extra
algebra up front but you can just go straightforward with
the power rule and you're going to get this exact same thing. So the whole takeaway here is
pause look at your expression. See if there's a way to
simplify it and it's especially a good thing if you can get
out of using the quotient rule 'cause that sometimes is
just hard to know or remember and even when you do remember
it, it can get quite hairy quite fast.