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uberclay

159 Posts

 Posted - 10/13/2007 :  01:50:34 I just finished a lesson on tangents which was capped off with the Power Rule and the Chain Rule for derivatives.I cannot figure out how to differentiate the following, given the rules of this lesson. Can someone please point in the right direction.I'm really unsure about this one. In all of the practice questions I was able to eliminate the denominator.Q: differentiate y = (8x^4 - 5x^2 - 2) / (4x^3)A: y = (8x^4 / 4x^3) - (5x^2 / 4x^3) - (2 / 4x^3)y = 2x - (5/4)x^-1 - (2/4)x^-3y' = 2 + (5/4)x^-2 + (3/2)x^-4Q: differentiate y = (5x)^(1/2) - (x/5)^(1/2)A:Similar problem with this one. None of the practice questions had coefficients in the radicand.I've split this one into two parts, (5x)^(1/2) and (x/5)^(1/2). for s = (5x)^(1/2)let t = 5x, so that s = t^(1/2)dt/dx = 5ds/dt = (1/2)t^(-1/2)chain rule: ds/dx = dt/dx * ds/dt, t = 5xtherefore ds/dx = 5 * (1/2)(5x)^(-1/2)ds/dx = 5 / 2(5x)^(1/2)for u = (x/5)^(1/2)let v = x/5, so that u = v^(1/2)dv/dx = 1/5du/dv = (1/2)v^(-1/2)chain rule: du/dx = dv/dx * du/dv, u = (x/5)^(1/2)thereforedu/dx = (1/5) * (1/2)[(x/5)^(1/2)](-1/2)du/dx = 5^(1/2) / 10x^(1/2)TF: dy/dx = [5 / 2(5x)^(1/2)] - [5^(1/2) / 10x^(1/2)]I am very sure that both these answers are way off base.Thanks for your time. Edited by - uberclay on 10/14/2007 01:28:11

skeeter

USA
5634 Posts

 Posted - 10/13/2007 :  07:32:50 first one is fine ... no need for the chain rule (why use it when you don't have to?)second one is fine, although you took a long road to arrive your derivative.y = (5x)1/2 - (x/5)1/2dy/dx = (1/2)(5x)-1/2*5 - (1/2)(x/5)-1/2*(1/5)dy/dx = 5/[2(5x)1/2] - 51/2/(10x1/2)which can be simplified further ...dy/dx = (2/5)(5/x)

uberclay

159 Posts

 Posted - 10/15/2007 :  23:13:43 Thank you for checking those for me Skeeter.My instructor expects me to take the long way, at times, to show a complete understanding of the concepts in the lesson. I milled over that second question tonight and did it w/o the Chain Rule, using that common factor David exposed. This time I got a different looking derivative but soon found that they were equal.David, I see what you are saying about joining at the finish in question 1. I'll take care of that tomorrow.g'night.

uberclay

159 Posts

 Posted - 10/16/2007 :  15:32:11 Is dy/dx = (8x^4 + 5x^2 + 6) / 4x^4a suitable answer for question 1?Thanks

skeeter

USA
5634 Posts

 Posted - 10/16/2007 :  18:42:44 yes
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