Problem

Find $ y' $ and $ y". $ $ y = \frac {1}{(1 + \ta…

00:30

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Problem 47 Hard Difficulty

Find $y^{\prime}$ and $y^{\prime \prime}$
$$
y=\cos (\sin 3 \theta)
$$

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Video Transcript

all right, we're going to find the first and second derivatives of this problem, and this is a composite function that has three layers. So the very inside layer is three theta. The middle layer is the sine function on the outermost layer. Is the coastline functions, so we'll be using the chain rule extended. So let's start by finding why prime the first derivative. So the derivative of the outside would be the derivative of co sign, and that is negative. Sign. We've negative sign of sign of three data. Now we take the derivative of the next layer and the next layer, remember was signed and the derivative of Sinus co sign. So we have the co sign of three data. Now we take the derivative of the next layer, and the next layer is three data, and it's derivative is three. Okay, we can clean this up a little bit. We have y prime equals negative three times a sign of the sign of three Seita times, a co sign of three data. We could have this part after the coastline part. It doesn't really matter, so that's the first derivative. Now let's find the second derivative So we're taking the derivative of the derivative. And what we have here is a product. So we're going to need to use the product rule and then inside the product rule, we're going to have the chain rule, and I'm going to tack on the negative three as part of the first term. So here we have the first term in the product and here we have the second term in the product. Okay, So the first times the derivative of the second plus the second times, the derivative of the first, that would be the product rule. So the first would be negative. Three. Sign of sign of three data times The derivative of the second derivative of co sign is negative sign. So times negative sign of three data times a derivative of the inside three. So that's the first times the derivative of the second, plus the second co sign of three data times the derivative of the first. Okay, so now we're doing the derivative of this whole thing, and we're going to use the chain rule. And again, it's a triple layer chain. So we have the negative three times the derivative of sign Co sign of sign of three data times the derivative of the next layer. The next layer would be the Inter sign, and it's derivative would be co signed. So coastline of three data times the derivative of the next layer and that would be three data and it's derivative is three. Okay, so we have the first times the derivative of the second, plus the second times, the derivative of the first. Now let's see what we can do to simplify it. So, looking at our first term, we have a negative three and a negative and a three. We can multiply all those together and we have nine. So we have nine sign of three theta times, a sign of the sign of three data. Now, for the other part, we have a negative three and we have a three, so that would be minus nine. So we've minus nine and we have co sign of three data and another co sign of three data so we could write that as co sine squared of three data times. The remaining part is co sign of sign three data. We can see just how much more complicated it gets when we take a second derivative

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Problem

Find $ y' $ and $ y". $ $ y = \frac {1}{(1 + \ta…

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kristen Karbon

Michael Jacobsen

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

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kristen Karbon

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Kristen Karbon

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals