dy/dx

[Guest]

I am trying to find dy/dx, if
ln x = cos(y)

Thanks

 

[tkhunny]

I am trying to find dy/dx, if
ln x = cos(y)

Implicit Differentiation
(1/x) = -sin(y)*(dy/dx)

 

[soroban]

Hello, ceerh!

I am trying to find \(\displaystyle \frac{dy}{dx}\;\) if: \(\displaystyle \ln x\:=\:\cos(y)\)

I assume this is Implicit Differentiation.

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

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

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


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It can be done explicitly, too . . .

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

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

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