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Express the function w(c) VTIthe form w(T) (Jf o g)(r) where f(z) + Iand glr) + f(r)61%-)Check AIl PrLS...

Question

Express the function w(c) VTIthe form w(T) (Jf o g)(r) where f(z) + Iand glr) + f(r)61%-)Check AIl PrLS

Express the function w(c) VTI the form w(T) (Jf o g)(r) where f(z) + Iand glr) + f(r) 61%-) Check AIl PrLS



Answers

In Exercises $61-68,$ compute the derivative indicated. $$ R(u, v, w)=\frac{u}{v+w}, \quad R_{u v w} $$

Okay. Don't get all scared because there's a hyperbolic sine there even easier than regular season and go sign because you don't have to remember where the plus or the minus sign goes. All right. We're gonna take the derivative with respect to you. You derivative of the hyperbolic sine hyperbolic co sign of the stuff times the derivative of the stuff. The derivative of the stuff with respect to you. So tha tha squares derivative is zero derivative of U. V. Is V. All right. If you you derivative of the hyperbolic coastline. Hyperbolic sine of the stuff times the derivative of the stuff and then times the V. That was already there. So, we have V square and hyperbolic sine. U. V plus data square. Now, I'm gonna take the derivative with respect to theta be scores constant, derivative of the hyperbolic sine. Hyperbolic co sign U. V plus data squared times the derivative of the stuff. U. And V are constants with respect to theta. So it's attributed to zero plus the derivative of theta squared. Uh huh. Truth data. All right.

Section three, about six problems. 16 derivatives that have to do with the chain room. What were asked to do here is to rewrite this function. Why as a composite function. So if I can figure out why as a function of you and then if I could figure out you, uh, as a function of X, then I could use the chain rule for its derivative. So what if we live f of you be coat engine of you and G of X Equal pi minus one over X, then what you see here? This is nothing more than f of G of X. Therefore, it's derivative. It's f prime of G of X times g, prime of X. So let's just go figure out these driven And so what is the derivative of f With respect to you, the druid of the co tangent. It's minus the coast seeking squared of you. And then you substitute the value of G into that and you get minus Khost Seacon squared. Hi, minus one over x. And then when you look at G, you take the derivative of see that's just one over X squared. So are derivative. Will be the product of these two quantities, so it's going to be one over X squared and so minus one over X squared and then the coast Seacon squared a pie minus one over X.

F. Of U. V. Is the co sign of U. Plus V. Squared. And so they want us to find the partial of F. With respect to you. You ve All right. So we're gonna do F. You F. Sub. You first of U. V. So the outside function is the co sign. The derivative of the coastline is minus the sign of the stuff. Times of those derivative of this stuff. And the derivative with respect to you. So that's just gonna be one because V. Is constant. I cannot take the derivative again with respect to you. The outside function is minus sign. So the derivative is minus the co sign of the stuff Times the derivative of this stuff which is one again can now take the derivative with respect to V. The outside function is minus the co sign. So it will be positive. The sign of the stuff times the derivative of the stuff

Okay. I'm gonna find the derivative with respect to you. You then V. Or in whatever order I want. So first I check it out, it's the co sign. So it's going to repeat back and forth. Cosine and sine because it's got a U. And V. In there. So the order I do it in is not going to matter. Someone take the derivative with respect to you. First derivative of the co sign of something is minus the sign of the something times the derivative of the something which is one uh F. U. U. The a derivative of minus the sign of something is minus the co sign of the something times the derivative of the something. All right now I'm gonna take the derivative with respect to be derivative of minus the coastline is sign of something times the derivative of the something. And that's the answer.


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