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(10 points)a. Assuming a,b and k are constants; calculate the following derivative_ ake^(kt) #([:] - = bke^(kt)b. Find a value of k so thatis a solution to &# ...

Question

(10 points)a. Assuming a,b and k are constants; calculate the following derivative_ ake^(kt) #([:] - = bke^(kt)b. Find a value of k so thatis a solution to &# 5 k =Find a value of k so thatis a solution to #&d. Write down the general solution in the form €1(t) =? and x2(t) =?, i.e-, write down a formula for each component of the solution: Use A and B to denote arbitrary constants. 81(t) E2(t)

(10 points) a. Assuming a,b and k are constants; calculate the following derivative_ ake^(kt) #([:] - = bke^(kt) b. Find a value of k so that is a solution to & # 5 k = Find a value of k so that is a solution to # & d. Write down the general solution in the form €1(t) =? and x2(t) =?, i.e-, write down a formula for each component of the solution: Use A and B to denote arbitrary constants. 81(t) E2(t)



Answers

Find the derivative. Assume $a, b, c, k$ are constants. $$y=5$$

Hi there and this problem, whereas to find the derivative of F of X equals. We're told before all these problems, by the way, that a be and C and K artist Constance So X is the only variable here. Notice we definitely have a quotient. So we're going to use the quotient rule. We'll call our top function f Actually, since we're already using F, we'll call our top function H Let's say caller bottom function G kind of like the book does. And so the quotient rule says that the derivative will be the bottom function times that river to the top, minus top times the derivative of the bottom all over the bottom squared. So all we have to do now is plugging our own top and bottom for H and G SOCOG. The bottom in our case is C X plus came many times the derivative of the top. The top is a X plus B minus. Now we need the top, which is a X plus be times the derivative of the bottom. The bottom is C express cape and then on the bottom you just get G squared. So time to take the derivatives on top. The bottom is not gonna change. So I'm just gonna put that in there right now. Okay, so on top this first parentheses, this is going to stay there. So next we're going to need this derivative here and the derivative of X plus B. No, I am. Beard is Constance. So the derivative of eight Times X is just a and the derivative of this plus B zero. Cosby's just a constant. So are derivative is actually just a That wasn't too bad. Now we have minus this. Parentheses stays as it is, and we just need now the derivative of CX plus que But this is exactly the same cases before the derivative of C C Times X plus Que Let's just see okay, So time just to simplify the top a little bit weaken Distribute. So we get Let's see, a time see Times X plus eight times k. All that just came from distributing this, eh? Minus gonna be a little careful here. We're going to distribute the minus azi, distribute the sea so we'll get minus. Get a time, see times X and minus B times c. Make sure you're convinced about that we distribute the Sian and then me to distribute the minus across Take it. And on bottom we just have everything the same as it was looking at the top, we noticed that we have a plus a c x and a minus a C X. So these actually cancel out And so all we're left with at the end you get eight times k minus b times c all over, see X plus que scored and that is our derivative hopefully.

Either in this problem, whereas to fund derivative of y equals five x times e to the X squared. The important thing to know here is that we've got one function of x times, another function of X. So we're definitely going to need the product rule. And so the derivative it will be, well, the first part. His five X Now we need two multiplied by the derivative of the second part now, plus the derivative of the first part times the second part. So that's what the product rule says. Now let's just take these derivatives. The derivative of either the X squared is either the X squared times. Now we need the chain rule. We need to multiply by the derivative of this inter function X squared, which is to X. And so that was the derivative eat of the X squared. Now, in the second term, the derivative of five X is just five. Then we keep the e to the X squared here. So we're basically done. We could just clean things up a little bit. Five times two is 10. We have x times. Another X will give us 10 x squared The X squared plus five feet of the X squared. And finally, we want to do even a little more. It's up to you. You can factor out that you to the X squared. We came from factor out of five as well, and that will leave us with Let's see, looks like two x squared close one in these parentheses.

If you watched it Turner before this one we're doing. But, uh uh, Constance, right? Yes. So this one is this because it's a fixed constant, right, Kay? Okay. He's a fixed constant so that they're gonna treat it the same way. This is gonna be f prime of X rate this two years. Gonna drop here to multiply the case. They're gonna have to k right. And then X now this to hear gonna take away 12 to minus one is one right to exit a part of oneness still likes. So this becomes too derivative, right for ah, it face Constant, Kate. Once again, Kate.

Either in this problem, were asked to find the derivative of y equals X times natural log of X. So since we have one function of X multiplied by a second function of X, we're definitely going to need the product rule. So the derivative will be Let's just follow the product rule. It's the first. The two factors times the derivative of the second factor, plus the derivative of the first factor times the second factor. Okay, it's the extra stays like that. The derivative of Log X, we know is one over X. The derivative of X is just one. So here we go. We're gonna simplify a little bit extended one over X gives us one in the second term. Isjust love X and we are done. Hopefully that helps


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