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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?

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

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

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

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  • $\begingroup$ Thank you for your reply! I almost understand but one question. How did you get (dx/dx)y turn into y? $\endgroup$ – Kijimu7 Feb 24 '20 at 22:45
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    $\begingroup$ @Kijimu7 We have $\frac{dx}{dx}=1$, i.e. the derivative of $f(x):=x$ with respect to $x$ is just the constant function $1$. $\endgroup$ – Dave Feb 24 '20 at 22:53
  • $\begingroup$ @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? $\endgroup$ – Kijimu7 Feb 24 '20 at 23:01
  • $\begingroup$ @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 $\frac{dy}{dx}$. $\endgroup$ – Dave Feb 24 '20 at 23:37
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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 $\frac{1-y}{x}=\frac{1-\frac{x-4}{x}}{x}=\frac{\frac{4}{x}}{x}=\frac{4}{x^2}$

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

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