# Using implicit differentiation on trigonometric fraction and logarithm

I am supposed to solve the following using implicit differentiation.

$$\sin\left(\frac{x}{y}\right)+\ln(y)=xy$$

This is what I have so far:

$$\cos\left(\frac{x}{y}\right)\frac{d}{dx}\left(\frac{x}{y}\right)+\frac{d}{dx}\big(\ln(y)\big)=y+x\frac{dy}{dx}$$

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.

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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

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

$$\cos\left(\frac{x}{y}\right)\frac{d}{dx}\left(\frac{x}{y}\right)+\frac{d}{dx}\big(\ln(y)\big)=y+x\frac{dy}{dx}\;.\tag{1}$$

You have

$$\frac{d}{dx}\left(\frac{x}y\right)=\frac{y\cdot1-x\frac{dy}{dx}}{y^2}=\frac1y-\frac{x}{y^2}\frac{dy}{dx}$$

and

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

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

$$\cos\left(\frac{x}y\right)\left(\frac1y-\frac{x}{y^2}y'\right)+\frac{y'}y=y+xy'\;,$$

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

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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.

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$$\sin(x/y) + \ln y = xy$$

$$\frac{\cos(x/y)(y-xy')}{y^2} + \frac{y'}{y} =xy' + y$$

$$\frac{y\cos(x/y) - xy'\cos(x/y)}{y^2} + \frac{y'}{y} = (xy' + y)\mid ~~\times ~~y^2$$

we have

$$y\cos(x/y) - xy'\cos(x/y) + yy' = xy^2y' + y^3$$

$$yy' - xy'\cos(x/y) - xy^2y' =y^3 - y\cos(x/y)$$

$$y'(y-x\cos(x/y)-xy^2)=y^3 - y\cos(x/y)$$

$$y'= \frac{y^3-y\cos(x/y)}{y-x\cos(x/y)-xy^2}$$

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