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7. For constants A and B. consider the function defined by f(At)? 48 t < 2 h(t) Bts t2 2 Find all values A and B such that h(t) is differentiable:...

Question

7. For constants A and B. consider the function defined by f(At)? 48 t < 2 h(t) Bts t2 2 Find all values A and B such that h(t) is differentiable:

7. For constants A and B. consider the function defined by f(At)? 48 t < 2 h(t) Bts t2 2 Find all values A and B such that h(t) is differentiable:



Answers

$72-74$ Assume that all of the functions are twice differentiable
and the second derivatives are never $0 .$
$$\begin{array}{l}{\text { (a) If } f \text { and } g \text { are concave upward on } 1, \text { show that } f+g \text { is }} \\ {\text { concave upward on I. }} \\ {\text { (b) If } f \text { is positive and concave upward on I, show that the }} \\ {\text { function } g(x)=[f(x)]^{2} \text { is concave upward on } 1 .}\end{array}$$

It's close when you reign here, so for have to be differential everywhere. It has to be differential. X equals to us. Well, so the derivative one approaching from both sides of X equals two has to be equal to one another. Sorry. When we differentiate half a pecs is equal to two X and we get four. Well, we know that M X plus be is a line. So that means that the slope has to be equal to four since that day since the differentiated equation is equal to em for X, this bigger been too. And now we're going to find for be so f of two is equal to four if X is less than or equal to two. So we're gonna solve for B. Why is he equal to M X plus be? We plug in four equal to four times two plus B and we gotta be value of negative for

So given the abuser for XY and XX g f d n y is a choppy we'll find by media time equals to do we need two seconds. Okay, so using changeover DW by DT now w is function of x and y and which is f so right this is there, left by their legs and then we have two different shape eggs which is function of P so well, right, the x or ditty. But w is also function of y. So we'll have another towns of delivered by dell y times, the white by people. So now we just have to substitute the values. So let's calculate the values first. So now if you see we need delivered by Dylex value when t equals to two. So we need to check What is the value of F when these equals to two Since we are given value of four Hama three So X is equals two g of two and gov. Two is nothing but four. So access for and why is edge of to from here and why it's nothing very expert Dr is nothing but please. So why is three so for this meaning f X or SOCOM, a tree which is given to us as minus five. So let's substitute it to minus five and DX by DT. It is nothing but G dash of T and G dash of two is minus one. Similarly, they left by Del y will be seven and d y by day, which is nothing much address of two, which is six. So this gives us five plus 42. It is required to 47 that is a solution. Thank you.

In problem three. We have this equation X wise equal to four, and we're gonna be doing some implicit differentiation, which is getting us prepared for the related rates problems that we're going to be doing it later in this section. Now, we're told that Wine X are functions of T, which means that we're gonna be having to employ the chain rule because we don't just have X and why we actually have x of t and y of tea if you wanted to think them as functions and rights in that way. But the first step is I wanted to go ahead and separate these two variables, so I'm actually gonna divide each side by X that we get Why is gonna be equal to four over X? I may be going. Wait a second, Caleb, you can't just divide by a variable because you may be dividing by zero. But in both cases, in part A and B for dealing with non zero values of X. So it's perfectly okay to be doing this. I'm going to go ahead and end work with this little bit more. I'm actually gonna make it equal to four times X to the negative one, so I find that a little bit easier than dealing with variable in the denominator, just turning it into a power. Now let's go ahead and we're prepared to take the derivative they have the derivative of why, with respect to t going to be equal to four now is using the power rules. So you bring that very, ah, that power out in front. So we have a negative one IMEs X and then we're going to subtract one from the power so that we have negative too. That's going to be multiplied by the derivative of X. With respect to t, we took the derivative of this function with respect to X. Then we have to take the derivative X with respect to t to complete our chain rules. Now, just to simplify this, we have that b y over d t is going to be equal to negative four over X squared times DX over DT Now that I have this general form, I could begin solving for the things I was asked to solve for So let's begin with art. A So, in part, a we need to solve for d y over DT. We're told that X is eight. So we have negative 4/8 squared and then that's gonna be multiplied my DX over DT, which is going to be 10. That's just gonna be negative. 4/64 times 10. I can simplify this fraction. I just going to get one in the numerator and then the nominator, I'll get 16 and I can go a little bit further. If I divide by two, I can see that I have five. This 10 turns into five and 16 will go into eight. That's just going to end up being negative 58 So that should be the answer. Maybe didn't quite follow how I simplified the fraction, but should be negative 5/8 now, going into part B, we're actually gonna be solving for DX over DT. They were given that Do Y over DT is equal to negative six. And then that's going to be equal to my negative for over my ex value squared, which is just one that's gonna be multiplied by my DX over DT. So we know that DX over duty it's gonna be equal to negative six over negative four, which is gonna be ABS

This video covers partial derivatives and the chain rule. In this problem, you are given that our S. And T. Is equal to achieve of U. And V. Where U. And V are. In terms of snt. You are also given the values of certain functions and partial derivatives at certain points to find the partial derivative of our with respect to us. We use the chain rule given here. We can set this equal to The partial derivative of our with respect to us at 12 is equal to the partial derivative of U. G. With respect to you. That you have one too The of 12 times the partial derivative of U. With respect to us. That one too, plus the partial derivative of G with respective V At. u of 12 V. of 1, 2 times the partial derivative of the with respect to S. At 12 Substituting in using the values given we get that this is equal to the partial derivative of G with respect to you at 57 times the partial derivative of U. With respect to ask that one too. What's the partial derivative of U. With respective V. At 57 times the partial derivative of V. With respect to S. At 12 Again substituting in This gets us nine times 4 Plus -2 times two Or 36 -4. Giving us our final answer. The partial derivative of our with respect to s at 12 is equal to 32. Okay. The partial derivative of our with respect to T at this point is also given by the chain role and this is equal to mhm. The partial derivative of our with respect to cheat at 12, it's equal to the partial derivative of G with respect to you At the same point as above. So 5 7 again times the partial derivative of U. With respect to T. At 12 plus the partial derivative of G. With respect to be again at the same point as above. 5, 7 times the partial derivative of U with respect to T. At 12 This substituting in is unequal to nine times negative three plus -2 times six for 27 -27 -12. Yeah, giving us our final answer. The partial derivative of our with respect to T. At 12 is equal to negative 39.


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