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Find +h first anJ Second Jerivative of ony Tuncton AefineJ imphcitly by Sx? +97-*...

Question

Find +h first anJ Second Jerivative of ony Tuncton AefineJ imphcitly by Sx? +97-*

Find +h first anJ Second Jerivative of ony Tuncton AefineJ imphcitly by Sx? +97-*



Answers

Find $A^{-1}.$ $$A=\left[\begin{array}{lr} e^{t} \sin 2 t & -e^{-t} \cos 2 t \\ e^{t} \cos 2 t & e^{-t} \sin 2 t \end{array}\right]$$

Asked ICO job the square bless fight day minus one. Give anybody square and we noticed that we can write down this function is that he's going over the square. That's fine. Day on with the square minus one over the square and you make Is it you do the riveter? We were good. Everything in Judah single exponents therefore gonna one from the first term plus five D about minus one for the second term. And the last time will be t bermanis too. And recall not, you have the expo and the re with Do which again, the and temps extra and minus one a blind this formula was You get now, the aspirin we coach you on sort of our father Constant cd riveting. Which again, is there here, therefore one ISO for this one minus five D by minister for this one on Blessed to debate one street for this. So that's when we the answer for the S Bram And for the second, the river till again Echo Joe for this one wish again. Ah 10 Derail ministry and finished only gonna minus six thing about when it's far. So that would be the answer for the secondary with you

For this problem. We're going to be finding first derivatives. So let's start by considering the derivative of FFT which is going to be T squared plus one cubed. So what we're gonna do is we're going to multiply the three in front giving us P squared plus one squared. And then using the chain road we're going to multiply this by treaty. So that would be our final answer. And then for the next one We're going to have the derivative with respect to pee. So that's going to give us six p. Using the power rule minus one half. Then for the next one it's going to be the derivative with respect to S So that's going to be one over to root S squared plus one. But by the chain role we need to multiply this by eight to S. So we do that. The truth cancel. And we're just left with he s on top that would be the first derivative there. Then we have first derivative with respect to T. So that's going to be a to a squared T plus B squared. Um That's it. And then we're going to find lastly the first derivative with respect to pee. So T squared even though it's a variable, it's just a constant when we're differentiating with respect to P. Such as going to give a zero plus um rather since we're taking the third power it's going to be three times T squared. Class, repeat. Not going to be squared now. And then we have to multiply this using the chain rule by the derivative. What's inside. So that's just going to be three, meaning that will happen nine right here, and that will be the final answer.

It's a little background info. Before we start this question, I won't be using the product rule in the chain rule. The product rule in respect to her or with respect to our question says that we pull the constant our routes. We have our times, the derivative with respect to our of the natural log a rhythm of r squared plus squared. We're going to add that to the natural log rhythm of r squared plus squared times the derivative with respect to our r Then we're also gonna be using the chain rule with respect to our so are derivative with respect to our In this case of the natural log rhythm of R squared plus squared is equal to the derivative of the outside. So one over r squared plus squared times the driven of the inside, which with respect to our gives us to our remember during this lesson, we're gonna be looking at Drew. It is with respect to certain variables. So any time we're looking at that, another variable will be considered a constant. So with respect to R s is considered a constant, which of course, the derivative, then a zero. So getting into the question. We have the product rule that we stand before so are times natural or sorry. The partial derivative, uh, the natural longer of them are squared plus desperate plus the natural longer them of r squared plus s murder times You're effective with respect to our are as we stated in the chain rule. With the first part of this question, we're going to have our times the derivative of the natural log rhythm of r squared plus squared. So remember that was one over r squared plus x squared times, the inside stories in chain rule. So too are because very finding the derivative with respect to our so that s becomes a constant. So the derivative a zero plus what we have left the natural order of them are smart. What's this? Where times the derivative are with her Steptoe are which is just one. So our first partial derivative with respect to our ends up being too are square or r squared plus desperate plus the natural logarithms are square. Plus I swear that's the first partial derivative with respect to our now we have to find the first partial derivative with respect to s with respect. It s we follow very similar steps. So again, using a product rule, we find that we have our times the derivative. Now remember, with respect s this time of the natural order of them are square plus yes, where plus the natural order of them Our square Plus s where times are with respect to us. So the dream of are with respect s writing that down and using a chain rule again. Or we will come to the chain role we have our times, the derivative with respect It s of the natural log a rhythm. So times won over our square plus ass spurt times to s Remember, this time we're taking the derivative with respect to s. So our becomes a constant and the derivative of a constant is Europe. So we add that to natural law algorithm, uh, are square less ness, word times the derivative of our with respect to s, which is zero. So our second partial derivative that we get in this question, I guess it's a first partial derivative. It's the 2nd 1 we're finding. So with respect, it s we get two R s over our swear s crypt. We've now solved the problem in which we found both first partial derivatives

Together we were given a function and we're asked to find the first push through bits of dysfunction function is R. P Q equals the inverse tangent of PQ squared. Someone told me a very perfect primary. Now to find the partial derivative or with respect to pee, I'll hold queue constant and differentiate with respect to T. So we get 1/1 plus P times Q squared squared by the chain rule, multiplied by Q squared. So this is Q squared over one plus P squared cubed before. Likewise, the popularity of our suspect to Q. Can you found by holding P constant and differentiate with respect to queue using the gene rule. This is 1/1 plus P times Q squared squared times the derivative of the inside with respect to Q. Which is two P. Q. And this is equal to two times P times Q over one plus P squared times Q. To the fourth. You some more. This creek melts into


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