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## Calculus 1

### Course: Calculus 1 > Unit 3

Lesson 2: More chain rule practice- Derivative of aˣ (for any positive base a)
- Derivative of logₐx (for any positive base a≠1)
- Derivatives of aˣ and logₐx
- Worked example: Derivative of 7^(x²-x) using the chain rule
- Worked example: Derivative of log₄(x²+x) using the chain rule
- Worked example: Derivative of sec(3π/2-x) using the chain rule
- Worked example: Derivative of ∜(x³+4x²+7) using the chain rule
- Chain rule capstone
- Proving the chain rule
- Derivative rules review

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# Worked example: Derivative of 7^(x²-x) using the chain rule

AP.CALC:

FUN‑3 (EU)

, FUN‑3.C (LO)

, FUN‑3.C.1 (EK)

Sal differentiates the exponential function 7^(x²-x) using our knowledge of the derivative of aˣ and the chain rule.

## Want to join the conversation?

- Can't we just differentiate without using the chain rule, I mean just using the a^x method of differentiating.for instance, ln[7]*[7]^x^2-x(8 votes)
- No because we must also multiply by the derivative of the inner function as well, which in this case is (x^2-x). So the derivative is (ln[7]*[7]^(x^2-x))(2x-1).(5 votes)

- At1:26in the video, Sal indicates that dy/dx is equal to dv/du ∙ du/dx. It seems that it should be dy/du rather than dv/du; such that dy/dx = dy/du ∙ du/dx; where you can think of the du’s cancelling leaving dy/dx. I know it is not cancelling, but it seems that in all the problems I’ve seen in the past, this is the case. Am I reading too much into the notation?

Or could it be that since y = v(u(x)) that dy/dx = dv/dx = dv/du ∙ du/dx?(9 votes) - I know that there is proof for the chain rule, but why can it both be applied to parenthesis as a product, and a power? They are totally separate, and have different order operations?(5 votes)
- The Chain Rule is used where you have a function of a function, in the form of f (g (x)). This is not the same thing as a product of functions. When we have the product of two functions, in the form f (x)*g (x), we use the Product Rule: f'(x)*g (x) + f (x)*g'(x). When we have the composite function f (g (x)), we use the Chain Rule: f '(g (x)*g'(x).

An example of a composite function would be e^sin (x), whose derivative is e^sin (x)*cos (x)(5 votes)

- Is there a way to rewrite 7^(x^2-x) ?(3 votes)
- You can apply the two properties: a^m/a^n = a^(m-n) and (a^m)^n = a^(mn) but it may or may not be any simpler.

7^(x^2-x) = 7^(x^2) / 7^x = (7^x)^x/7^x

or

7^(x^2-x) = 7^(x(x-1)) = (7^x)^(x-1) = (7^(x-1))^x(5 votes)

- Why are we using the ln at all? Wouldn't just the chain rule work? and the answer could be 7^(x^2-x) (2) ? How do you know to use the ln in the first place?(2 votes)
- Check out earlier videos. When you have a^x you can not use power rule, instead you derive it as a^(x) * ln (a)

I do not know at all where you came up with your answer.(5 votes)

- At1:31, Sal wrote dv/du times du/dx. Shouldn't it be dy/du times du/dx?(2 votes)
- Yes, this is a mistake, dy/dx should be equal to dy/du * du/dx which becomes

[d(7^(x^2-x)) / d(x^2-x)] * [d(x^2-x) / dx]

= [7^(x^2-x) * ln(7)] * [2x - 1]

You can clearly see why it should be this way, as we evaluate the first term the whole power of 7 can be treated as a single variable that's why the form is same as ln(a)*a^x (notice we didn't do anything else for the first term)

If it were dv/dx then it wouldn't make sense, it would look like

d(7^x) / d(x^2-x)

where we have no function 7^x.(2 votes)

- v'(x)=ln(7)*7^x. How is that? I know it's an identity, but I'd like to see proof of this identity.(2 votes)
- Do you need to use the chain rule twice here? Once to find the derivative of the outer function and then again to find the derivative of y?(2 votes)
- You only need to apply once. Given that e^x d/dx = e^x you need to make a substitution so you can get in that form.(2 votes)

- If we rewrote 7^(x^2-x) as 7^(x^2)/7^x and differentiated using the chain rule and then the quotient rule, would we get the same answer?(2 votes)
- When using the chain rule, must you first fully simplify the derivative of the "inner" function before multiplying it with the derivative of the "outer" function? Or is this not necessary?(2 votes)

## Video transcript

- [Voiceover] Let's say
that y is equal to seven to the x squared minus x power. What is the derivative
of y, derivative of y, with respect to x? And like always, pause this video and see if you can figure it out. Well, based on how this has
been color-coded ahead of time, you might immediately recognize that this is a composite function,
or it could be viewed as a composite function. If you had a v of x, which
if you had a function v of x, which is equal to seven to the xth power, and you had another function u of x, u of x which is equal
to x squared minus x, then what we have right over here, y, y is equal to seven to something, so it's equal to v of,
and it's not just v of x, it's v of u of x, instead of an x here you have the whole function u of x, x squared minus x. So, it's v of u of x and
the chain rule tells us that the derivative of
y with respect to x, and you'll see different notations here, sometimes you'll see it
written as the derivative of v with respect to
u, so v prime of u of x times the derivative
of u with respect to x, so that's one way you could do it, or you could say that this is equal to, this is equal to the
derivative, the derivative of v with respect to x, sorry,
derivative of v with respect to u, d v d u times the derivative
of u with respect to x, derivative of u with respect to x, and so either way we can
apply that right over here. So, what's the derivative
of v with respect to u? What is v prime of u of x? Well, we know, we know,
let me actually write it right over here, if v of x is
equal to seven to the x power v prime of x would be equal
to, and we've proved this in other videos where we
take derivatives exponentials of bases other than e, this
going to be the natural log of seven times seven to the x power. So, if we are taking v prime of u of x, then notice instead of an x everywhere, we're going to have a u of x everywhere. So, this right over
here, this is going to be natural log of seven times seven to the, instead of saying seven to the x power, remember we're taking v prime of u of x, so it's going to be seven to
the x squared minus x power, x squared, x squared minus x power, and then we want to multiply
that times the derivative of u with respect to x. So, u prime of x, well, that's going to be two x to the first which
is just two x minus one, so we're going to
multiply this times two x, two x minus one, so there you have it, that is the derivative
of y with respect to x. You could, we could try to simplify this or I guess re-express
it in different ways, but the main thing to realize is, look, we're just gonna take the
derivative of the seven to the this to the u of x
power with respect to u of x. So, we treat the u of x the
way that we would've treated an x right over here, so it's
gonna be natural log of seven times seven to the u of x power, we take that and multiply
that times u prime of x, and once again this is just an application of the chain rule.