Flying Shogi Posted November 10, 2012 Share Posted November 10, 2012 (edited) First up is simplifying: The question asks to find the derivative of the function f(x)=x^{2}(x-2)^{4} 1. I got f'(x)=(x^{2})4(x-2)^{3}(1)+(x-2)^{4}(2x) How do you get it to look like this-> 2x(x-2)^{3 }(3x-2) Please don't tell me I'll have to FOIL/use Pascal's triangle and then factor 2. y=(1/4)(sin2x)^{2 } I have y'=sin2xcos2x How do you make it looks like-> y'=(1/2)(sin4x) 3. Then it's finding the derivative of a function at a given point on a graphing calculator The function is g(t)= (3x^{2})/root(t^{2}+2t-1) @ (1/2,3/2) How would you put this in the calculator? I put (3x^{2})/root(x^{2}+2x-1) in Y_{1 }and then I went to MATH, then pressed 8 and put this: nDerv((3x^{2})/(root(x^{2}+2x-1))X,1.5) in Y_{2} I got an error message when I tried to graph it. It says "ERR:Argument" Next is using implicit differentiation to find dy/dx 4.How do you do this? x^{1/2}+y^{1/2}=9 The answer is -root(y/x) 5. x^{3}-xy+y^{2}=4 I got (-3x^{2})/(2y-x) but the book says it's (y-3x^{2})/(2y-x). I don't understand where that y in the numerator is coming from. Lastly is h(x)=sin2xcos2x To find the derivative, what should be done first? Do the Chain Rule first or Product Rule first? Or do the Product RUle as it is and do the Chain rule when I get to the dv and du terms? I'm kinda confused. Edited November 10, 2012 by MagicLugh Quote Link to comment Share on other sites More sharing options...

Balcerzak Posted November 10, 2012 Share Posted November 10, 2012 (edited) First up is simplifying: The question asks to find the derivative of the function f(x)=x^{2}(x-2)^{4} 1. I got f'(x)=(x^{2})4(x-2)^{3}(1)+(x-2)^{4}(2x) How do you get it to look like this-> 2x(x-2)^{3 }(3x-2) Please don't tell me I'll have to FOIL/use Pascal's triangle and then factor Factoring out the common factor of (x-2)^{3 } leaves 4x^{2} + (x-2)(2x) or 4x^{2} + 2x^{2} -4x or 6x^{2} - 4x or 2x(3x-2) 2. y=(1/4)(sin2x)^{2 } I have y'=sin2xcos2x How do you make it looks like-> y'=(1/2)(sin4x) Trig double/half angle identities. Learn them. 3. Then it's finding the derivative of a function at a given point on a graphing calculator The function is g(t)= (3x^{2})/root(t^{2}+2t-1) @ (1/2,3/2) How would you put this in the calculator? I put (3x^{2})/root(x^{2}+2x-1) in Y_{1 }and then I went to MATH, then pressed 8 and put this: nDerv((3x^{2})/(root(x^{2}+2x-1))X,1.5) in Y_{2} I got an error message when I tried to graph it. It says "ERR:Argument" No clue what calculator you're using. Try reading the manual. 4.How do you do this? x^{1/2}+y^{1/2}=9 The answer is -root(y/x) (1/2)x^{-1/2} dx + (1/2)y^{-1/2} dy = 0 or y^{-1/2} dy = -x^{-1/2} dx or dy/dx = -x^{-1/2} / y^{-1/2} or dy/dx = - (x/y)^{-1/2} or dy/dx = - (y/x)^{1/2} 5. x^{3}-xy+y^{2}=4 I got (-3x^{2})/(2y-x) but the book says it's (y-3x^{2})/(2y-x). I don't understand where that y in the numerator is coming from. What are we supposed to be solving for here? 3x^{2}dx - ydx -xdy + 2ydy = 0 or 3x^{2}dx - ydx = xdy - 2ydy or dy (x-2y) = dx (3x^{2}-y) or dy/dx = (3x^{2}-y) / (x-2y) Lastly is h(x)=sin2xcos2x To find the derivative, what should be done first? Do the Chain Rule first or Product Rule first? Or do the Product RUle as it is and do the Chain rule when I get to the dv and du terms? I'm kinda confused. I have no idea why you're talking about dv and du. Sounds like you're trying to integrate by parts or something. Stop that. This shouldn't be giving you problems. Also, IMO the first step should probably be simplifying using Trig Identities, tbqh. Edited November 10, 2012 by Balcerzak Quote Link to comment Share on other sites More sharing options...

Grace Posted November 10, 2012 Share Posted November 10, 2012 (edited) Also, IMO the first step should probably be simplifying using Trig Identities, tbqh. 2sin(x)cos(x) = sin2x use that, should help I'll fiddle with the rest myself but Bal pretty much hit the rest on the head edit: confirming pretty much everything Bal said. I did it myself and got to the same answers Edited November 10, 2012 by Manix Quote Link to comment Share on other sites More sharing options...

Caliban of Sycorax Posted November 10, 2012 Share Posted November 10, 2012 My calc teacher in high school described the last question as a "bubble" of sorts... don't you derive what's inside the trig identity then derive the identity? And then the derivative of the inside is the coefficient of the whole equation? If I'm right then it should be -4cos(2x)sin(2x) Quote Link to comment Share on other sites More sharing options...

Grace Posted November 10, 2012 Share Posted November 10, 2012 you can simplify the cos(2x)sin(2x) using 2sin(x)cos(x) = sin(2x) ie sin(x)cos(x) = sin(2x)/2 => sin(2x)cos(2x) = sin(4x)/2 sin(4x)dx = 4cos(4x) => (sin(4x)/2)dx = 2cos4x and then simplify if possible Quote Link to comment Share on other sites More sharing options...

Flying Shogi Posted November 10, 2012 Author Share Posted November 10, 2012 (edited) It's actually was supposed to say AB Calc but somehow it became AM Factoring out the common factor of (x-2)^{3 } leaves 4x^{2} + (x-2)(2x) or 4x^{2} + 2x^{2} -4x or 6x^{2} - 4x or 2x(3x-2) Can't believe I didn't see that. (1/2)x^{-1/2} dx + (1/2)y^{-1/2} dy = 0 or y^{-1/2} dy = -x^{-1/2} dx or dy/dx = -x^{-1/2} / y^{-1/2} or dy/dx = - (x/y)^{-1/2} or dy/dx = - (y/x)^{1/2} For this one, since we derive usually with respect to x, I thought there is no need for dx. Instead, I was taught that a dy/dx should be written every time a y is derived. Edited November 10, 2012 by MagicLugh Quote Link to comment Share on other sites More sharing options...

Integrity Posted November 10, 2012 Share Posted November 10, 2012 There is always a need for a dx. Quote Link to comment Share on other sites More sharing options...

Balcerzak Posted November 10, 2012 Share Posted November 10, 2012 For this one, since we derive usually with respect to x, I thought there is no need for dx. Instead, I was taught that a dy/dx should be written every time a y is derived. You told me to do implicit differentiation. That is how you do implicit differentiation. (At least it was when I learned how to.) You differentiate every variable available, using the chain rule as necessary, and leave behind the d's. Quote Link to comment Share on other sites More sharing options...

Flying Shogi Posted November 11, 2012 Author Share Posted November 11, 2012 (edited) You differentiate every variable available, using the chain rule as necessary, and leave behind the d's. I understand this but the way I learned it, I wasn't required to put a dx every time a variable with a x is derived. Instead, I was taught to just write dy/dx when a y is derived. Then move things to the other side and factor as needed. Either way, I eventually got same answer. Edited November 12, 2012 by MagicLugh Quote Link to comment Share on other sites More sharing options...

Balcerzak Posted November 11, 2012 Share Posted November 11, 2012 Sounds like only cosmetic differences then. I think my way is more intuitive, especially when it comes time to Diff. Eq., but that's probably just my biases. Quote Link to comment Share on other sites More sharing options...

Flying Shogi Posted November 12, 2012 Author Share Posted November 12, 2012 So I just got to the Related Rates section of the homework and I'm screwed. This is the first question: Find the rate of change of the distance between the origin and a moving point on the graph y=x^{2}+1if dy/dt=2 cm/sec. and I'm drawing a blank. I got to this: dy/dt=2x(dx/dt) dy/dt=4x Then I'm lost. I feel like I'm doing this wrong. The answer is [2(2x^{3}+3x]/root(x^{4}+3x^{2}+1) Any hints? Quote Link to comment Share on other sites More sharing options...

shroudening Posted November 12, 2012 Share Posted November 12, 2012 (edited) You have two equations, believe it or not! Not only do you have y = x^2 + 1, but you also have d = sqrt(x^2 + y^2) by the Pythagorean theorem. Think about it. You're not trying to find the change in distance of x or y, but for both. So since you know that dy/dt = 4x and dx/dt = 2, you can implicitly derive dd/dt (the change in distance over time), which is what you are trying to find. Once you find dd/dt, substitute (x^2+1) into y, so that your answer contains only x's. Hopefully that helps... I did the problem out and got the correct answer. Edited November 12, 2012 by Isaac55 Quote Link to comment Share on other sites More sharing options...

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