## 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?

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 Quote by studentxlol 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.

 Recognitions: Gold Member Science Advisor Staff Emeritus 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?