I need to find dy/dx $\displaystyle y^2=lnx$ Thanks Jason
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Hello there, Simply differentiate both sides, and then isolate $\displaystyle \frac{dy}{dx} $. Good luck!
Last edited by scherz0; Jun 17th 2009 at 11:52 AM. Reason: Thanks, HallsofIvy.
Originally Posted by scherz0 Hello there, Simply differentiate both sides, and then isolate $\displaystyle \frac{dy}{dx} $. Remember that $\displaystyle \frac{dy}{dx} \ln x = \frac{1}{x} dx $. No. The $\displaystyle \frac{d\ln x}{dx}= \frac{1}{x}$. What $\displaystyle \frac{dy}{dx} \ln x$ is depends on what function y is of x. Good luck!
Originally Posted by Darkhrse99 I need to find dy/dx $\displaystyle y^2=lnx$ Thanks Jason You know, I hope, that the derivative of $\displaystyle y^2$ with respect to y is 2y. Use the chain rule to find the derivative of [itex]y^2[/itex] with respect to x. Set that equal to the derivative of ln x with respect to x and solve for dy/dx.
hi your answer should be equal to : $\displaystyle \frac{dy}{dx} = \frac{1}{2\,x\,\sqrt{log\left( x\right) }}$
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