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APPENDIX B TABLE OFDERIVATIVES General Formulas dc) =0 duf6) + gkr)) = f' (x) + g' (r) df(r)g(r)) = f' (r)g(x) + f(r)g' (r) (product rule) dr&qu...

Question

APPENDIX B TABLE OFDERIVATIVES General Formulas dc) =0 duf6) + gkr)) = f' (x) + g' (r) df(r)g(r)) = f' (r)g(x) + f(r)g' (r) (product rule) dr") =nr"- for real numbers n dcfkr)) = cf' (x) du6r) - g(r)) = f' (x) - g' (r) 43) = E()f' () = f(x)g' (2 (quotient rule) (g(r)i? dflg(r)ll = f" (gtr)} > (chain rule) Trigonometric Functions dsinx) cosX Ztanr) sec? x(secx) scrtanx(cosx) ~sinxdcotr) a(cscx) ~CeCrCO

APPENDIX B TABLE OF DERIVATIVES General Formulas dc) =0 duf6) + gkr)) = f' (x) + g' (r) df(r)g(r)) = f' (r)g(x) + f(r)g' (r) (product rule) dr") =nr"- for real numbers n dcfkr)) = cf' (x) du6r) - g(r)) = f' (x) - g' (r) 43) = E()f' () = f(x)g' (2 (quotient rule) (g(r)i? dflg(r)ll = f" (gtr)} > (chain rule) Trigonometric Functions dsinx) cosX Ztanr) sec? x (secx) scrtanx (cosx) ~sinx dcotr) a(cscx) ~CeCrCO



Answers

Derivatives of other trigonometric functions Verify the following derivative formulas using the Quotient Rule. $$\frac{d}{d x}(\sec x)=\sec x \tan x$$

For this question. They give us co Seacon, and they want us to prove the formula for its derivative. Using the quotient rule. Well, what is co seeking its one over Sign? The Rex? Let's do our question role. We'll take the derivative of the top, multiplied the bottom but doesn't matter because it's speech 70 times it. And then we'll subtract the derivative of the bottom times the top all over the bottom squared and we end up with coastline over negative co sign over Science Court. What is that? We have a negative Cosi connects and a coat engine x One of the power is saying convey be pulled apart legs and a coastline X over sign Next could be left separately From there you go.

We're ants to take the derivative of this function two ways. Once as a quotient and once as a chain we're going to use the quotient rule. We do need to rewrite this function so that it is starting as it starts as a quotient. We have one over X plus syntax quotient role tells us the derivative would be the bottom times the derivative of the top which is zero minus the top Times the derivative of the bottom which is one plus cosign X. And that's all over the bottom square. What we should notice here is that the first two terms here are gonna be multiplied by zero. So and if we distribute the negative one, we end up with negative one minus cosign axe over X plus syntax quantity square. Now let's compare that to what we would get. Using the chain rule under the chain role. We would bring the power out front. We would subtract one from the old power so that's negative too. And then we multiply by the derivative, the inside which is one plus Cosign X. This can be rewritten, distribute the negative one into the one plus Cosign X. That's negative. One minus coast sign X. And then the X plus sign. Access to the negative second power so we can move it to the bottom. And we should see that we get the same answer.

This time they want us to verify that Riveted Attention, Rex. Using the Kocian. Well, it's here we can do. We have a co signer Rex over sign of Rex, and we're gonna take the derivative of that. So that's just the derivative of the first function. Negative sign of Rex Times of autumn so that it becomes a squared and then minus the derivative of the bottom times it up. So co sign square direct over sine squared MX. Well, that's ends up being a negative one because of our trade identity science where plus coastline squared is equal to one. And then what's one over sine squared that's just co seeking squared a negative coast He can't square Durex.

We are using the portion rule to prove that the derivative of co sequence of X is equal to negative co sequence coat inject. So first we want to identify be quotient rule F over G prime is equal to f prime times G minus G prime times f All of this is over g squared. We know that co Sequent of X is equal Teoh one over side of X f will be one mg Will these sign of X? So let's put it all together you will get d over DX of one multiplied by sine of X minus D over DX of side of X well supplied by one. All of this is over sine squared x So what we're going to do next is find are derivatives D over DX off one is equal to zero next D over DX of sign of X is equal to co sign X. Now that we have a derivatives, let's put everything back together. We will get zero times Sign of X minus coastline X well supplied by one All of this is over sine squared X simplified. We will get negative co sign of X over sine squared X. Our next step is to use this identity. Co sign of X over sign of X Was equal to CO attention to Next The application of this rule will look like negative co tangents of X over side of X. After this, let's use one last identity. One over side of X is equal to co secret texts, so the application of this identity will look like negative co tangent. X Co Sequent X. This is the final answer.


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