dy/dx

Hello, ceerh!

I am trying to find dydx  \displaystyle \frac{dy}{dx}\; if: lnx=cos(y)\displaystyle \ln x\:=\:\cos(y)
I assume this is Implicit Differentiation.

We have: cos(y)=ln(x)\displaystyle \:\cos(y)\:=\:\ln(x)

Then: sin(y)(dydx)=1x\displaystyle \:-\sin(y)\cdot\left(\frac{dy}{dx}\right)\:=\:\frac{1}{x}

Therefore: \(\displaystyle \L\:\frac{dy}{dx}\:=\:-\frac{1}{x\cdot\sin(y)}\)


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It can be done explicitly, too . . .

We have: cos(y)=ln(x)\displaystyle \:\cos(y)\:=\:\ln(x)

Then: y  =  cos1[ln(x)]\displaystyle \:y\;=\;\cos^{-1}[\ln(x)]

Therefore: \(\displaystyle \L\;\frac{dy}{dx}\:=\:\frac{-1}{\sqrt{1\,-\,[\ln(x)]^2}}\,\cdot\,\frac{1}{x} \;= \;\frac{-1}{x\sqrt{1\,-\,[\ln x]^2}}\)
 
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