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Find the x coordinate of the stationary point of the following curves |
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| Nov28-11, 04:24 AM | #1 |
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Find the x coordinate of the stationary point of the following curves
1. The problem statement, all variables and given/known data
Find dy/dx and determine the exact x coordinate of the stationary point for: (a) y=(4x^2+1)^5 (b) y=x^2/lnx 2. Relevant equations 3. The attempt at a solution (a) y=(4x^2+1)^5 dy/dx=40x(4x^2+1)^4 40x(4x^2+1)^4=0 Find x... How? (b) y=x^2/lnx dy/dx=2xlnx-x^2 1/x / (lnx)^2 2xlnx-x^2 1/x / (lnx)^2=0 Find x... How? |
| Nov28-11, 04:38 AM | #2 |
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Then either 40x = 0 or (4x^2+1)^4 = 0. and solve the above two equations. |
| Nov28-11, 06:58 AM | #3 |
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You are aware that [itex]x^2/x= x[/itex] aren't you?
[itex]y= x^2/ln(x)[/itex]: [itex]y'= (2xln(x)- x)/(ln(x))^2= 0[/itex] Use parentheses! What you wrote was [itex]y'= 2x ln(x)- (x/(ln(x))^2)= 0[/itex]. Multiply both sides of the equation by [itex](ln(x))^2[/itex] and you are left with 2x ln(x)- x= x(2ln(x)- 1)= 0. Can you solve that? |
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