# Using implicit differentiation, what is $\frac{dy}{dx}$ if $xy + 4 = x$?

Using implicit differentiation, what is $$\frac{dy}{dx}$$ if $$xy + 4 = x$$?

I have a few questions regarding the implicit differentiation problem above.

I know the answer is $$\frac{1 - y}{x}$$ but I don't know how to get the answer.

So, I believe it is using a product rule. However, do I also need to use the chain rule?

What I did is

$$xy + 4 = x$$ I got $$x + y$$ using the product rule so then,

$$x\left(\frac{dy}{dx}\right) + y + 0 = 1$$

And is $$4$$ a constant?

I am not sure the next step why $$\frac{dy}{dx}$$ belongs to $$x$$ instead of $$y$$? Can anyone please explains this to me?

• As $y=\frac{x-4}{x}$, you can express the derivative as a function of x and moreover you could have directly derived y. $\frac{dy}{dx}=\frac{4}{x^2}$. – Jean-Claude Colette Feb 24 '20 at 22:48

$$xy+4=x$$

Differentiate both sides with respect to $$x$$; the product rule says $$\frac{\mathrm d}{\mathrm dx}(xy)=\frac{\mathrm dx}{\mathrm dx}y+x\frac{\mathrm dy}{\mathrm dx}$$. So you get

$$y+x\frac{\mathrm dy}{\mathrm dx}+0=1\implies\frac{\mathrm dy}{\mathrm dx}=\frac{1-y}x$$

• Thank you for your reply! I almost understand but one question. How did you get (dx/dx)y turn into y? – Kijimu7 Feb 24 '20 at 22:45
• @Kijimu7 We have $\frac{dx}{dx}=1$, i.e. the derivative of $f(x):=x$ with respect to $x$ is just the constant function $1$. – Dave Feb 24 '20 at 22:53
• @Dave thank you for explaining that. I understand now! Thank you :) I have one more question. What does y respect to x mean in this case? – Kijimu7 Feb 24 '20 at 23:01
• @Kijimu7 "with respect to $x$" indicates that the differentiation is on the variable $x$. So "the derivative of $y$ with respect to $x$" really means $\frac{dy}{dx}$. – Dave Feb 24 '20 at 23:37

The above solutions explain the implicit differentiation well, just to make a complete case, here you can get the explicit formula of $$y$$ as a function of $$x$$ instead of using implicit differentiation.

$$xy+4=x \Rightarrow y=\frac{x-4}{x}\Rightarrow y'=\frac{dy}{dx}=\frac{(1)(x)-1(x-4)}{x^2}=\frac{4}{x^2}$$

Please notice that this is equal to $$\frac{1-y}{x}=\frac{1-\frac{x-4}{x}}{x}=\frac{\frac{4}{x}}{x}=\frac{4}{x^2}$$

$$xy +4=x \implies d/dx(xy+4) = d/dx(x) \implies y + x dy/dx = 1 \implies dy/dx = (1-y)/x$$