Find the x coordinate of the stationary point of the following curves

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SUMMARY

The discussion focuses on finding the x-coordinate of stationary points for two functions: (a) y=(4x^2+1)^5 and (b) y=x^2/lnx. For the first function, the derivative dy/dx is calculated as 40x(4x^2+1)^4, leading to the stationary point when 40x=0 or (4x^2+1)^4=0. For the second function, the derivative dy/dx simplifies to (2xlnx - x)/(lnx)^2, and the stationary point is found by solving 2xlnx - x = 0. The discussion emphasizes the importance of proper notation and algebraic manipulation in solving these equations.

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studentxlol
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Homework Statement



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

Homework Equations




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?
 
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studentxlol said:
40x(4x^2+1)^4=0

Find x... How?

re 1st prob:

Then either 40x = 0 or (4x^2+1)^4 = 0.

and solve the above two equations.
 
You are aware that x^2/x= x aren't you?

y= x^2/ln(x): y'= (2xln(x)- x)/(ln(x))^2= 0
Use parentheses! What you wrote was y'= 2x ln(x)- (x/(ln(x))^2)= 0.

Multiply both sides of the equation by (ln(x))^2
and you are left with 2x ln(x)- x= x(2ln(x)- 1)= 0. Can you solve that?
 

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