Find y′′ for the curve ln(x) + y = ln(x^2) − y^2 at y = 0?
Note that the domain for x is that x 0. There will only be one point on the curve with y = 0

Question

Find y′′ for the curve ln(x) + y = ln(x^2) − y^2 at y = 0?
Note that the domain for x is that x > 0. There will only be one point on the curve with y = 0

Christopher Agresta · Accepted Answer

#(d^2y)/(dx^2)= pm 3# #logx+y=logx^2-y^2rArrlog(xe^y)=log(x^2e^(-y^2))# Assuming #x > 0# we have #xe^y=x^2e^(-y^2)# or #e^(y+y^2)=x# or #y^2+y-log x=0# now solving for #y# #y = 1/2(-1pmsqrt(1+4logx))# which is real for #1+4logx ge 0# or #0 < x # and now #(d^2y)/(dx^2) = pm(3 + 4 Logx)/(x^2 (1 + 4 Logx)^(3/2))# Now for #e^(y+y^2)=x# we have #y=0 rArr x = 1# and then #(d^2y)/(dx^2)= pm 3#

Evelyn Agard · Answer

#(d^2y)/dx^2 = -(4y^2+4y+3)/(x^2(1+2y)^3) #

For #x=1# we have #y(1)=0# and #y''(1) = -3# Given the implicit function: #lnx +y = ln(x^2) -y^2# defined for #x > 0#, note that based on the properties of logarithms: #ln(x^2) = 2lnx# so that the equation becomes: #y^2+y = lnx# and then for #x=1# we have #y=0#. Differentiate both sides of the equation: #2ydy/dx +dy/dx = 1/x# #dy/dx (1+2y) = 1/x# and again: #(d^2y)/dx^2(1+2y) +2(dy/dx)^2 = -1/x^2# substituting: #dy/dx = 1/(x(1+2y) )# we have: #(d^2y)/dx^2(1+2y)= -2/(x(1+2y) )^2 -1/x^2# #(d^2y)/dx^2 = -1/(x^2(1+2y)) (2/(1+2y) ^2 +1)# #(d^2y)/dx^2 = -1/(x^2(1+2y)^3) (2+(1+2y) ^2 )# #(d^2y)/dx^2 = -1/(x^2(1+2y)^3) (2+1+4y+4y^2 )# #(d^2y)/dx^2 = -(4y^2+4y+3)/(x^2(1+2y)^3) # For #x=1#: #[(d^2y)/dx^2]_(x=1) = -3 #

Paisley Johnson · Answer

undefined To find $ y'' $ for the given curve at $ y = 0 $, we first need to find $ y' $ and then differentiate it again.

Given the equation $ \ln(x) + y = \ln(x^2) - y^2 $, differentiate both sides with respect to $ x $ to find $ y' $:

$$ \frac{d}{dx} (\ln(x) + y) = \frac{d}{dx} (\ln(x^2) - y^2) $$

$$ \frac{1}{x} + y' = \frac{2x}{x^2} - 2yy' $$

$$ y' = \frac{2x}{x^2} - 2yy' - \frac{1}{x} $$

Now, we substitute $ y = 0 $ into the equation:

$$ y' = \frac{2x}{x^2} - 2 \cdot 0 \cdot y' - \frac{1}{x} $$

$$ y' = \frac{2}{x} - \frac{1}{x} $$

$$ y' = \frac{1}{x} $$

Now, differentiate $ y' $ with respect to $ x $ to find $ y'' $:

$$ \frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2} $$

Thus, $ y'' = -\frac{1}{x^2} $.

Find y′′ for the curve ln(x) + y = ln(x^2) − y^2 at y = 0? Note that the domain for x is that x &gt; 0. There will only be one point on the curve with y = 0

Find y′′ for the curve ln(x) + y = ln(x^2) − y^2 at y = 0? Note that the domain for x is that x > 0. There will only be one point on the curve with y = 0