Using implicit differentiation on trigonometric fraction and logarithm

Question

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.

Answer 1

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 $y'$.

Answer 2

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.