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2) Find the third derivative of f(x) = e2 +3x+cosx 3) Find y' ify= csc(x)....

Question

2) Find the third derivative of f(x) = e2 +3x+cosx 3) Find y' ify= csc(x).

2) Find the third derivative of f(x) = e2 +3x+cosx 3) Find y' ify= csc(x).



Answers

Find the derivatives of the given functions. $$y=\frac{2 \csc 3 x}{x^{2}}$$

All right. In this question, we want to find a given derivative F two of X. That is the second derivative of Fx. For Activex is equal to three times two minus X squared cute! This question is challenging our understanding of higher order caritas in particular is testing our knowledge of the fact that for higher order derivatives, you only have to use the standard differentiation techniques. That is we use all the normal differentiation rules even before where we take the derivative one by one. What this means is that we have to take the derivative Fx once to obtain F one, then we have to take a second derivative of F one to obtain F two. So we take two driven sequentially from F to obtain F two. So taking F one, we first have F one equals three times three times two minutes, X squared square times negative two X. This apply the power rule to our two minus x squared cube and then the chain rule to the content of parentheses. That's our F one is negative 18 x two minus x squared square. Thus, by the product rule and chain rule again, we have F double prime is the following, which simplifies to negative 18 times two minutes X squared square plus 72 X squared times two minus x squared.

All right. We want to find dy dx for the function. Why is equal to four X cubed minus three coast. You can route to expose three. And this problem we're going to use derivative shortcuts to solve in particular. We're going to be relying on newfound econometric derivatives that we learned. So relevant rules are list this year. Trade derivatives are D V X and X equals co sets and so on. And particularly going to rely on D V X. Costa can't X is negative. Costa can export tangent X. We also make note of the chain rule, product rule and so on. The chain rule will be essential for this problem. So we can rewrite our problem as well equals four X cubed minus three. Costa can't to expose 3 to 1 half. This makes using the chain rule of it simpler. Thus we can solve. Dy dx is 12 X squared plus three coast. You can't to expose three. Co change to expose three. This is for the coast can't derivative by the chain rule. We multiply by half to expose three negative one half times to giving a solution. Dy dx equals 12 X squared plus three coast. You can route to expose three co. Change in route to expose three all over the route to expose three.

We want to find the derivative of the function of X equals two X cubed minus three, raised to the power of one third. So we're going to need to rely on some of the calculus rules that we picked up for differentiating these are shortcuts, which makes it easier than using the traditional definition of the derivative. So let's define those rules. Now we have one through three. The power rule one D D X. X. To the Z equals X. To the minus one to the product rule. DDX F G equals D F D, F G plus Suggs. And three of the chain rule DDX F M G F X equals F. Prime G. Of X times G. Prime X. So we can apply the chain rule and power rules to solve this problem. Traditionally we would rewrite F as a way that's easier to solve. But it is already in the simplest form. So that means you can go ahead and differentiate using shade and power rules so that's fine. Is one third some two X huge minus three to negative two thirds. That's by the power rule multiplying by six X squared is because of the chain rule That is six x graders derivative of two XQ -3. That's our solution is two X cubed or rather two X squared divided by two xq minus three. Raise the power of two thirds.

For this problem, we are asked to find the second derivative of the function F of X equals three times sine of X. So we start off by taking our first derivative. So it will be three times the derivative of sine of X, which will be CAS affects. So we get three CAs of X as our first derivative, and then we differentiate again. So it will be three times the derivative of Cossacks, which was Synnex. So we'll have negative three sine of X as our second derivative.


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