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## Strategy in differentiating functions

# Differentiating functions: Find the error

AP.CALC:

FUN‑3 (EU)

## Video transcript

- [Instructor] What we're gonna do in this video is look at
the work of other people as they try to take derivatives and see if their reasoning is correct and if it's not correct, try to identify what they should have done or where their reasoning went wrong. So over here it says Nate tried to find the derivative of X squared plus five X times sine of X. Here is his work. Is Nate's work correct? If not, what's his mistake? So pause the video and see
if you can answer this. Is Nate's work correct, and if not, what's his mistake? All right so I'm assuming
you've had a go at it. So let's work though this step by step. So over here he's just trying to apply the derivative operator to the expression. Which is exactly what he needed to do. He's trying to find the
derivative of this thing, and he says okay this is a
product of two expressions, and then he says okay,
well this is gonna be the same thing as the derivative or this is the same thing as
the product of the derivatives. Now this is a problem. You are probably familiar, if I take the derivative
of the sum of two things. So the derivative with
respect to X of F of X plus G of X, that indeed is equal to
the derivative of the first F prime of X plus the
derivative of the second, but that is not true if we
are dealing with the product of functions. The derivative with respect
to X of F of X, G of X is not necessarily, maybe there's some very
special circumstance, but in general it's not going to be just the product of the derivative. It's not going to be just
F prime of X, G prime of X. Here we would want to
apply the product rule. This is going to be equal to, this is going to be
equal to the derivative of the first function
times the second function, plus the derivative, or
let me write it this way. Plus the first function,
not taking its derivative, times the derivative
of the second function. So he should have applied
the product rule here, and so let's do that just to see what his
answer should have been. So what he should have done here, I'll get my correcting red pen out here. Say no that's not what
he should have done. He says let's take the
derivative of this first thing. So actually let me color code it. So the derivative of
this is two X plus five. So it should have been two X plus five times the second thing. So times sine of X. Times, we do it in another color. Time the sine of X, and then to that he would
add the first thing, which is X squared plus five X times the derivative of the second thing. So the derivative of
sine of X is cosine of X. So this is what we should have been seeing at this step right over here. He shouldn't have just taken the product of the derivatives. He should have applied the product rule. So his work is not correct, and his mistake is that he
didn't apply the product rule. He just assumed that the
derivative of the products is the same thing as the
product of the derivatives. Let's do more examples. Okay so let's see. It says Katy tried to find the derivative of two X squared minus four. All of that to the third power. Here's her work. Is Katy's work correct? If not, what is her mistake? So once again, pause the video. See if you can figure it out. All right now let's inspect Katy's work. So she's taking the derivative
of this and let's see, over here it looks like she's taking the derivative of the entire expression with respect to the inner expression and that is close to applying
the chain rule properly but it's not applying
the chain rule properly. So her work is not correct, and her mistake is she's not correctly applying the chain rule. Just as a review, the
chain rule says look, if we're trying tot ake the
derivative with respect to X of F of G of X, F of G of X, that this is going to be equal to, this is going to be
equal to the derivative of the whole thing with respect to G of X. So I could write that
as F prime of G of X. F prime of G of X times the
derivative of the inner function with respect to X. Times G prime of X. So over here, we could view our F function as a thing that takes its input and
takes it to the third power, and so this right over
here is F prime of G of X. So this thing is the F prime of G of X but she forgot to multiply
it by the derivative of the inner function with respect to X. So she forgot to multiply
this times the derivative of two X squared minus
four with respect to X. Which is gonna be, let's see, the derivative of two X squared, power rule two times two is four, so it's gonna be four X to the first, and then the derivative of
negative four is just a zero. So it's just gonna be times four X. So that's what she needed
to do in order for it to be correct. So she had to have this times four X here. Times four X. So not correct. She didn't correctly apply the chain rule. So let's do another one of these. So here, it says Njoman tried to
find the derivative of sine of seven X squared plus four X. Here's his work. Is Njoman's work correct? If not, what is his mistake? Pause the video, see if
you can figure it out. All right, so it's a derivative
of sine of this expression. So you'd wanna use the chain rule. In fact, using the chain
rule you wanna find the derivative of the outside function with respect to the inside. So the derivative of sine of something with respect to that something is gonna be cosine of that something. So that's right, and then you wanna multiply that times the derivative of the
inside with respect to X. So the derivative of
seven X squared is 14 X. The derivative of four X is four. So this is actually, that step looks good. But then Njoman does
something strange over here. This is the cosine of seven
X squared plus four X. Then that whole thing
times 14 X plus four, but they get confused where just looking at these parentheses and this
tends to happen sometimes. This is actually one of the key errors that the folks at the college board, AP folks told us about. Is that when dealing with
these transcendental functions cosine, sine, tangent, natural log that are written like this and people see the parentheses and see another parentheses, their brain just says oh let me multiply these two expressions in parentheses, but that's not right, because if we were to add parentheses this is what this is implying. So you can't just take the 14 X plus four and multiply it by this, and assuming you're taking
the cosine of the whole thing. So this is where Njoman makes the mistake. The work is not correct, and the mistake is trying to multiply these two expressions
and taking the cosine of the whole thing. Let's do one more of these. I find these strangely fun. All right this one is involved. Tom tried to find the derivative of the square root of X over X to the fourth. Here is his work. Is Tom's work correct? If not, what's his mistake? Pause the video and see if
you can figure that out. So it looks like he's trying
to apply the quotient rule. So applying the quotient rule, you would, in the numerator you
would take the derivative of the first expression
times the second expression and then minus the first expression times the derivative of
the second expression all of that over, or I should say the derivative
of the numerator expression times the denominator expression, minus the numerator expression
times the derivative of the denominator expression, all of that over the
denominator expression squared. So this looks correct actually. It's a correct application
of the the quotient rule. It looks like Tom is
correctly simplifying. So the derivative of X to the one half is one half X to the negative one half. So that looks right. Derivative of X to the fourth
is four X to the third. So that looks right. All of this looks algebraically right, and let's see, when you simplify this. So let's see, X to the negative one half times X to the fourth is indeed X to, well that's going to be X to the. Oh this correlates. So this simplifies to that. Which looks correct, and
that simplifies to that. Which looks correct. We're just using exponent
properties there, and then divide everything by, let's see, oh there, everything
is terms of X to 3.5. So we're going to have
negative 3.5 X to 3.5, and then you use exploder problems. So actually it looks like
he did everything correctly. This is the right answer. So his work is correct. He did not make any mistakes. But I do have a bone to pick. So speak with Tom, because he didn't have to
apply the quotient rule here. He did all of this hairy calculus and algebra but it could have been a very simple simplification he could have made up here, and this is a key thing to realize. He could have said hey, you know what, this is the same thing. It's the derivative with respect to X of X to the one half. That's what the square root of X is. Times X to the negative four power. That's what one over X to the fourth is. And so let me color code it. So that is the same thing as that, and that is the same thing as that, and you wouldn't even have
to use the product rule here. You could simplify this even further. This is the same thing as the derivative with respect to X of just, we have the same base. We can add the (mumbles) products. It's gonna be X to the negative
3., X to the negative 3.5, and so you can just use the power rule. So this is going to be equal to, bring the negative 3.5 out front. Negative 3.5 X to the, and then we just
decrement this one by one. We subtract one from that. Negative 4.5 power. So as you can see. He could have gotten this answer much much much much much much quicker, but he didn't make any mistakes. There's a little bit of a judgment error just immediately going
forth with the quotient rule which gets quite hairy quite fast.