ZiggyM Posted October 20, 2011 Report Share Posted October 20, 2011 How do you find the derivative of ln(2y+1)=xe^y??You need to use implicit differentiation for this, I do HL so I know it's required for that but not sure at the other levels.xe^y derives to: e^y + xe^y dy/dxdue to implicit differentiation and the chain rulethe derivative of ln(2y+1) = (2/(2y+1))(dy/dx) because the derivative of ln(f(x)) = f'(x)/(f(x)so the derived function of your function is (I rearranged it to = 0 )0 = e^y + xe^y dy/dx - (2/(2y+1))(dy/dx) The dy/dx is used whenever you differentiate y, so you use the chain rule. Does that make sense? If you don't know about implicit differentiation I'm not sure it will.. Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted October 20, 2011 Report Share Posted October 20, 2011 How do you find the derivative of ln(2y+1)=xe^y??You need to use implicit differentiation for this, I do HL so I know it's required for that but not sure at the other levels.xe^y derives to: e^y + xe^y dy/dxdue to implicit differentiation and the chain rulethe derivative of ln(2y+1) = (2/(2y+1))(dy/dx) because the derivative of ln(f(x)) = f'(x)/(f(x)so the derived function of your function is (I rearranged it to = 0 )0 = e^y + xe^y dy/dx - (2/(2y+1))(dy/dx) The dy/dx is used whenever you differentiate y, so you use the chain rule. Does that make sense? If you don't know about implicit differentiation I'm not sure it will..it makes sense but the answer says it is e^y(2y+1)/2-xe^y(2y+1) is that the same thing sorry i have no idea.... Reply Link to post Share on other sites More sharing options...
ZiggyM Posted October 20, 2011 Report Share Posted October 20, 2011 How do you find the derivative of ln(2y+1)=xe^y?? You need to use implicit differentiation for this, I do HL so I know it's required for that but not sure at the other levels. xe^y derives to: e^y + xe^y dy/dx due to implicit differentiation and the chain rule the derivative of ln(2y+1) = (2/(2y+1))(dy/dx) because the derivative of ln(f(x)) = f'(x)/(f(x) so the derived function of your function is (I rearranged it to = 0 ) 0 = e^y + xe^y dy/dx - (2/(2y+1))(dy/dx) The dy/dx is used whenever you differentiate y, so you use the chain rule. Does that make sense? If you don't know about implicit differentiation I'm not sure it will.. it makes sense but the answer says it is e^y(2y+1)/2-xe^y(2y+1) is that the same thing sorry i have no idea.... Try and rearrange it so that you have dy/dx = your terms, you might find that makes the answer make a bit more sense And check my differentiation too actually - I'm pretty brain dead atm Reply Link to post Share on other sites More sharing options...
dessskris Posted October 20, 2011 Author Report Share Posted October 20, 2011 (edited) find the angles AOB, AOC, BOC using cos rule, then find the area of each small triangle using sin rule. last, add them up.EDIT:find the angles AOB, AOC, BOC in terms of x using cos rule. since the three angles add up to 360°, I guess you can then find the values of each angle. then find the area of each small triangle using sin rule. last, add them up. Edited October 20, 2011 by Desy Glau Reply Link to post Share on other sites More sharing options...
dessskris Posted October 20, 2011 Author Report Share Posted October 20, 2011 question guys... if you're given the graph of f'(x) and are asked to draw f(x), how do you know the y coordinate of x? I think we can only find stationary and inflexion points, right? if they ask us to draw f(x) on the same axes as its derivative, how do you know if it should be taller or shorter than the derivative? in case if my question doesn't make sense, I'm asking about the vertical stretch of the function (not how to stretch it, but how stretched it is). I had some kind of mocks today and it came out in paper 2...dammitHow do you find the derivative of ln(2y+1)=xe^y??are you taking HL or SL? if you're taking HL, use implicit differentiation as Ziggy suggested. otherwise:ln(2y+1)=xe^yx=ln(2y+1)/e^yfind dx/dy from this. after that, you can find dy/dx from 1/(dx/dy). Reply Link to post Share on other sites More sharing options...
Lero Posted October 20, 2011 Report Share Posted October 20, 2011 That would be really difficult, I guess you could only estimate the y-coordinates depending on the slope of a tangent at that point. Like if the derivative graph is curving upwards during an interval then your f(x) also increases.Surely they cannot deduct marks if all the critical points are positioned correctly and the concavities are also correct? Reply Link to post Share on other sites More sharing options...
dessskris Posted October 20, 2011 Author Report Share Posted October 20, 2011 That would be really difficult, I guess you could only estimate the y-coordinates depending on the slope of a tangent at that point. Like if the derivative graph is curving upwards during an interval then your f(x) also increases. Surely they cannot deduct marks if all the critical points are positioned correctly and the concavities are also correct? ahhh I see... thank you! unfortunately I left that question to do last even though I've already worked out the stationary and inflexion points and the shape on a SCRAP PAPER.... in the end I didn't manage to copy the graph in the question booklet because I didn't realise the time was almost up oh well... Reply Link to post Share on other sites More sharing options...
iHubble Posted October 20, 2011 Report Share Posted October 20, 2011 I thought about this one too, but I was looking to avoid sin/cos/tan since I would like to solve it by hand... Reply Link to post Share on other sites More sharing options...
ZiggyM Posted October 20, 2011 Report Share Posted October 20, 2011 If it is supposed to be able to be solved by hand, then you can definitely still use sin cos and tan! You should know the values you can get using specific angles and their corresponding values. So if it is intended to be non-calculator, it is highly likely that it is going to be values like this because ... well just because Don't try and avoid using trig! You won't get far without it, especially in paper 1. Reply Link to post Share on other sites More sharing options...
jonathan810 Posted October 21, 2011 Report Share Posted October 21, 2011 (edited) Differentiate with respect to x:f(x)=ln [(x+23)/x) Edited October 21, 2011 by jonathan810 Reply Link to post Share on other sites More sharing options...
dessskris Posted October 21, 2011 Author Report Share Posted October 21, 2011 Differentiate with respect to x:f(x)=ln [(x+23)/x)2^3=8y=ln [(x+8)/x]y=ln (1 + 8/x)dy/dx= -8/x2(1 + 8/x) Reply Link to post Share on other sites More sharing options...
ZiggyM Posted October 22, 2011 Report Share Posted October 22, 2011 Hi ! My option for HL Maths is Sets, relations and groups. I'm generally fine with the questions that actually involve numbers (isn't that supposed to be what maths is!?!) but I often get very stuck on questions that are purely theoretical. So what I'm wondering is does anyone have any tips on how to get at least a few marks on these questions? I often find it simply hard to start the questions. I like Cayley tables and all that and I mostly have all the theory down, but it's when it comes down to applying this to purely theoretical situations that I come unstuck. Help? Btw, by theoretical I mean questions of the type: (a) Define an isomorphism between two groups (G, * ) and (H, •). (b) Let e and e' be the identity elements of groups G and H respectively. Let f be an isomorphism between these two groups. Prove that f (e) = e'. © Prove that an isomorphism maps a finite cyclic group onto another finite cyclic group. (Total 10 marks) Although not necessarily this question in particular. I have mark schemes, but I just find it difficult to understand them! From what I've gathered this isn't the most common option topic, but any help would be appreciated Thanks! Reply Link to post Share on other sites More sharing options...
Muhammad Fiqri Posted October 23, 2011 Report Share Posted October 23, 2011 Hi, just wanna ask,why does the graph of square root of 25 minus x-square didn't touch the x-axis, thus making it suspended?thank you for your time to answer this Reply Link to post Share on other sites More sharing options...
ZiggyM Posted October 23, 2011 Report Share Posted October 23, 2011 Could you perhaps be a bit clearer?Do you mean SQRT25 - x^2orSQRT(25-x^2) ? Reply Link to post Share on other sites More sharing options...
kinguncaged1 Posted October 23, 2011 Report Share Posted October 23, 2011 (edited) Well, the equation y=sqrt(25-x^2) is that of a circle centered around the origin with a radius of 5. If we rearrange the equation we get y^2+x^2=25 an equation for a circle. on a graphing calculator however it will only give positive values of y so that it is a function. so essentially, to get the real graph, you would have to graph both the positive and negative y's simultaneously. AKA: y=sqrt(25-x^2) and y=-sqrt(25-x^2) Edited October 23, 2011 by kinguncaged1 1 Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted October 23, 2011 Report Share Posted October 23, 2011 how do you find m: For what values of m is the line y=mx+5 tangent to the parabola y=4-x^2 Reply Link to post Share on other sites More sharing options...
CkyBlue Posted October 23, 2011 Report Share Posted October 23, 2011 (edited) Hello Bieber fan. Derivative of y=4-x^2 is -2x which is the formula that gives slope.and then you acquaint -2x to the line.y=(-2x)x+5if you want to find the tangent line, the line shares a common point with the parabola. therefore you acquaint the line and the parabola.(-2x)x+5=-x^2+4Solve for x.x^2=1x=+/- 1going back to the derivative, the derivative is then-2x .. and when you sub the x values, the slope(m) is either 2, or -2Check your work.2x+5=-x^2+4x^2+2x+1=0x=-1-2x+5=-x^2+4x^2-2x+1=0x=1 Edited October 24, 2011 by Capt'n Marth did the wrong question 1 Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted October 24, 2011 Report Share Posted October 24, 2011 (edited) Why thank you for solving the problem...I do not think i would've thought of that and yes I am a huge bieber fan Last problem for the night.. the line y=16x-9 is a tangent to the curve y=2x^3+ax^2+bx-9 at the point (1,7). Find the values of a and b. Please help Edited October 24, 2011 by hellokitty818 Reply Link to post Share on other sites More sharing options...
Lero Posted October 24, 2011 Report Share Posted October 24, 2011 Basically if the line y = 16x - 9 is tangent to the curve, then at the point (1,7) they have the same slope (derivative). dy/dx = 6x^(2) + 2ax + b. Now plug the x-coordinate given into this derivative and you should get 2a + b + 6 as an expression for the slope at the given point.Now since y = 16x - 9 is linear then the slope is always 16. So equate 2a + b + 6 = 16, therefore 2a + b = 10.Now to find a second equation relating the unknowns simply plug in your given co-ordinates into the cubic equation. So, 7 = 2(1) + a(1) + b(1) - 9. Therefore a + b = 14.Now you have two equations relating your two unknowns. Solving them simultaneously should give you a = -4 and b = 18. 1 Reply Link to post Share on other sites More sharing options...
Lero Posted October 24, 2011 Report Share Posted October 24, 2011 The easiest way to determine the points of the tangents concerning quadratic equations lies in the use of the discriminant. You know when b^(2) - 4ac = 0 then the equation contains only a single root? Apply the same logic here. If a line is tangent to a quadratic curve then it can meet it only once. So if you solve the equations:3x^(2) - x + 4 = mx + 1. Rearrange and you will get 3x^(2) - (m + 1)x + 3 = 0.Use the discriminant b^(2) - 4ac = 0 to determine the points of m which are tangent to f(x). Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.