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Find the derivative of the function. $ y = 2^{3^…

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Problem 43 Hard Difficulty

Find the derivative of the function.
$ g(x) = (2ra^{rx} + n)^P $

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October 10, 2019

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March 4, 2019

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Video Transcript

Okay, here we have a chain rule problem, and it looks like there was a shortage of numbers for this problem. And so somebody had to write it with a whole bunch of letters. But we're going to use all the same concepts and just apply all the rules we already know. So to find the derivative, the outermost layer is the P power. And so we're going to bring down the P and raise the inside to the P minus one, just like if that was a three or a four, you'd bring down that number and you'd raise it to the one less power. Now we need to multiply by. The derivative of the Insight and the insight is a sum of two terms. So we'll find the derivative of the first term, plus the derivative of the second term. So now we're finding the derivative of to our times a to the Rx, and so too are is just a constant are that are variable. Here is X. So all the other later letters are constants. So we leave. The two are the constant and we multiply it by the derivative of a to the Power Rx well, remember the derivative of a to the X is age the X Times natural law, gay. So if it's a to the Rx using the chain rule, the derivative would be a to the Rx Times natural log a times a derivative of our X, which is our So we get a to the Rx times natural law. Gay times are so that takes care of that. And then we have plus the derivative of n But an is just a constant. So it's driven of a zero. So we're actually not adding anything to that. Now let's see if there's anything we can do to this to make it a little bit nicer. Niedere Well, I suppose we could multiply the P and the two are and they are together. So we have to r squared P, and then we have the natural log of a as well. And then we have the big quantity here. We'll just rewrite it to our age of the R X plus n to the P minus one. And we still have the A to the Rx two, right, So a lot of it was just copying the same thing all over again.

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Watch More Solved Questions in Chapter 3

Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Samuel Hannah

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Samuel Hannah

James StewartCalculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Samuel Hannah

James Stewart

Calculus: Early Transcendentals