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

In this worked example we explore the process of differentiating the exponential function 7^(x²-x). We Leverage our previous understanding of the derivative of aˣ and the chain rule to unravel the complexities of this composite function to find its derivative.

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• 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
• 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).
• 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?
• 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)
• At in 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?
• Is there a way to rewrite 7^(x^2-x) ?
• 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
• 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?
• 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.
• At , Sal wrote dv/du times du/dx. Shouldn't it be dy/du times du/dx?
• No, you can write dy/du but when you do that, you don't know which y you should take. With this notation, it is clear you should take v. When you write dy/ du, the y = 7^(x^2 - x). But you have to take the v and not the y of the whole function in this case.
• 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.