Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am supposed to solve the following using implicit differentiation.


This is what I have so far:


My problem here is differentiating the fraction as well as the $\ln y$, especially since $\ln y$ does not contain $x$ to differentiate towards.

Thanks for your help in understanding this.

share|improve this question
Judging by the right hand side, $y$ appears to be a function of $x$. Then you can also use the chain rule on the left hand side, for example $\frac{d}{dx} \ln y = \frac{1}{y} \frac{dy}{dx}$. –  Eepzy Aug 12 '12 at 9:46

3 Answers 3

up vote 2 down vote accepted

Keep doing what you did on the right-hand side to get


You have



$$\frac{d}{dx}\ln y=\frac1y\frac{dy}{dx}\;;$$

both of these are just applications of the chain rule (together with the quotient rule and the rule for differentiating the natural log). Now for convenience I’ll write $y'$ for $dy/dx$; then $(1)$ becomes


and all that remains is to solve for $y'$.

share|improve this answer
Thank you, this makes sense a lot, but there is something i want to point out. Your result of the derivative of the fraction still needs to be multiplied by cos(x/y), which unfortunately then makes this a bit more complicated. But thank you for the clear calculations! –  Michael Frey Aug 12 '12 at 10:05
@Michael: Gah. That was really awful: I also managed to louse up the right-hand side by reading the wrong line, but it’s fixed now. Thanks for catching it. –  Brian M. Scott Aug 12 '12 at 10:10

Try to just grit your teeth and apply the rules exactly as you know them in these problems. On your last issue, note that $\ln(y)$ does contain an $x$, when you consider that $y=y(x)$ is a function of $x$, so we'll be able to use the chain rule.

Specifically, you know the derivative of $\log(f(x))=\frac{f'(x)}{f(x)}$. So, when $f$ is $y$,...

As to the fraction, again just apply the quotient rule. Denominator times derivative of numerator minus numerator times derivative of denominator-you know all those things, right?-over denominator squared.

share|improve this answer
 sin (x/y) + lny = xy
  cos (x/y)(y-xy')/y^2 + y'/y =xy' + y
  ((ycos(x/y) - xy'cos(x/y))/y^2 + y'/y =xy' + y)(y^2)
  we have;
  ycos(x/y) - xy'cos(x/y) +  yy' = xy^2y' + y^3
  yy' - xy'cos(x/y) - xy^2y' =y^3 - ycos(x/y)
 y'(y-xcos(x/y)-xy^2)=y^3 - ycos(x/y)  
share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.