I am trying to find dy/dx, if ln x = cos(y) Thanks
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,339 Mar 28, 2006 #2 ceerh said: I am trying to find dy/dx, if ln x = cos(y) Click to expand... Implicit Differentiation (1/x) = -sin(y)*(dy/dx)
ceerh said: I am trying to find dy/dx, if ln x = cos(y) Click to expand... Implicit Differentiation (1/x) = -sin(y)*(dy/dx)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Mar 28, 2006 #3 Hello, ceerh! I am trying to find dydx \displaystyle \frac{dy}{dx}\;dxdy if: lnx = cos(y)\displaystyle \ln x\:=\:\cos(y)lnx=cos(y) Click to expand... I assume this is Implicit Differentiation. We have: cos(y) = ln(x)\displaystyle \:\cos(y)\:=\:\ln(x)cos(y)=ln(x) Then: −sin(y)⋅(dydx) = 1x\displaystyle \:-\sin(y)\cdot\left(\frac{dy}{dx}\right)\:=\:\frac{1}{x}−sin(y)⋅(dxdy)=x1 Therefore: \(\displaystyle \L\:\frac{dy}{dx}\:=\:-\frac{1}{x\cdot\sin(y)}\) ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ It can be done explicitly, too . . . We have: cos(y) = ln(x)\displaystyle \:\cos(y)\:=\:\ln(x)cos(y)=ln(x) Then: y = cos−1[ln(x)]\displaystyle \:y\;=\;\cos^{-1}[\ln(x)]y=cos−1[ln(x)] Therefore: \(\displaystyle \L\;\frac{dy}{dx}\:=\:\frac{-1}{\sqrt{1\,-\,[\ln(x)]^2}}\,\cdot\,\frac{1}{x} \;= \;\frac{-1}{x\sqrt{1\,-\,[\ln x]^2}}\)
Hello, ceerh! I am trying to find dydx \displaystyle \frac{dy}{dx}\;dxdy if: lnx = cos(y)\displaystyle \ln x\:=\:\cos(y)lnx=cos(y) Click to expand... I assume this is Implicit Differentiation. We have: cos(y) = ln(x)\displaystyle \:\cos(y)\:=\:\ln(x)cos(y)=ln(x) Then: −sin(y)⋅(dydx) = 1x\displaystyle \:-\sin(y)\cdot\left(\frac{dy}{dx}\right)\:=\:\frac{1}{x}−sin(y)⋅(dxdy)=x1 Therefore: \(\displaystyle \L\:\frac{dy}{dx}\:=\:-\frac{1}{x\cdot\sin(y)}\) ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ It can be done explicitly, too . . . We have: cos(y) = ln(x)\displaystyle \:\cos(y)\:=\:\ln(x)cos(y)=ln(x) Then: y = cos−1[ln(x)]\displaystyle \:y\;=\;\cos^{-1}[\ln(x)]y=cos−1[ln(x)] Therefore: \(\displaystyle \L\;\frac{dy}{dx}\:=\:\frac{-1}{\sqrt{1\,-\,[\ln(x)]^2}}\,\cdot\,\frac{1}{x} \;= \;\frac{-1}{x\sqrt{1\,-\,[\ln x]^2}}\)