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No CakuktarDifferentiate the following with respect to 2: '((99z) ) + el + sin (3r)(b) Hence find: COF (+i '((99r)*) sin (r)...

Question

No CakuktarDifferentiate the following with respect to 2: '((99z) ) + el + sin (3r)(b) Hence find: COF (+i '((99r)*) sin (r)

No Cakuktar Differentiate the following with respect to 2: '((99z) ) + el + sin (3r) (b) Hence find: COF (+i '((99r)*) sin (r)



Answers

Find the indicated derivative. $$ r=\frac{x y^{2}}{z^{3}} ; \quad x=\cos s, y=\sin s, z=\tan s ; \quad \frac{d r}{d s} $$

It's clear. So when you married here. So we're gonna apply the square of a binomial, get nine a square. Class six A plus one. We're gonna different sheet both sides and we get nine to a to the two minus one, which is the power rule class six. This becomes zero, since it's the constant, it's 18 a plus six.

Uh make sure you pay close attention to what you're trying to find Sign of two Theta. Because in this problem is the first one where you're encountering that there is something special for your implicit differentiation. If at any point you're confused, um I would encourage you to pay attention to this thing. What's in the denominator because that's your independent variable. So when you go to take the derivative of the left side, that's just your power rule. You're taking the derivative of R. You have to write out D. R. D. Say to. Whereas on the right side, I'm assuming, you know that the derivative of sine is co sign you leave that to thought alone. This is technically the changing times, the derivative of two theta, which would just be too so uh you know, at some point you might you might be confused on what is your independent variable because they does your independent variable. You don't have to write D theta. D. Theta. Um So just to finish this problem to solve for D. R. D. Theta is a I guess first I'll write out that I'm dividing both sides by two because I'm not going to write that as part of my final answer because the two is will cancel their um But you do also need to divide this are over. So your final answer be co sign of truth. They divided by our mm. Mhm. Mhm.

Section 3.6 number 45 We're dealing with derivative that requires to know the chain role. So we're asked to find the derivative our prime, as a function of Fada. So this is gonna be the product rule. You've got two different functions multiplied. So what we want to do first is let's take the derivative of the first term. So the derivative of sine squared. So what's the derivative sign is the coastline. So this is gonna be the co sign of data squared Now, what is the derivative of Data Square? That's to thinner times. The co sign of two things was product rules. So this is you think that this is f and G, and this is F prime G plus G prime F. So plus the derivative of the second term. So the derivative of the co sign is minus sign of tooth Ada, the derivative of Tooth Ada is going to be too times the sign of data squared and the trouble just a little bit of organization that can go on here. So final answer r squared of Fada is going to be equal to toothache. The co sign of they the squared the co sign of tooth data and then minus two sign of data squared sign of two data and that is your final answer.

Yeah, we need to find the derivative. Do I D X. For the function? Why is equal to X divided by the arc sine of three X. To do so we're going to use derivatives shortcuts. We picked up through a single variable calculus in particular the inverse trig and metric derivatives we recently learned. So we have the inverse trig derivatives here. We're gonna need to use the DDX arc sine X equals 1/1 minus X. Where to solve. We're also gonna make use of the chain will product rule first. We can rewrite Y as Y. X times arc sine three X to the negative first. This allows us to start the derivative the product rule. So we then have dy dx equals one over arc sine three X. This is taking the river of x minus X over arc sine three X squared Times three over Route one Panasonic Square is the rib of are designed to extend a good first for the general simplifying we have you I D x equals square one minus nine X squared arc sine three x minus three X. All over Route one minus nine X squared times arc sine three X squared.


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