Implicit Derivative of $2xy=0$
I'd always thought it was solved like this:
$$ \frac{d}{dx}(2xy)=0=\frac{d}{dx}(x\cdot2y)= 2y$$ as $\frac{d}{dx}(kx) = k$
However according to multiple online differentiators, the actual answer is:
$$2xy′+ 2y =0$$
-- I realized the product rule - $$\frac{d}{dx}(ab) = a\cdot b′ + a′ \cdot b $$ - gives the results from the websites but why doesn't the rule $$\frac{d}{dx}(kx) = k$$ apply?
$\endgroup$ 13 Answers
$\begingroup$Because $y$ depends also of $x$, $y=y(x)$ so you have the product rule.
$x'=1$ and $y(x)'=y'$.
The rule that you said it is only valid if $y(x)$ were constant.
$\endgroup$ $\begingroup$You have:
$$\frac{\mathrm{d}y}{\mathrm{d}{x}} (2xy) = 0$$
Here are few things you need to remember:
- $x$ is a variable and not a constant.
- $y$ is a function of $x$ i.e. $y(x)$
The rule that you mentioned only works if you are multiplying a variable to a constant. In this case, you are multiplying a variable to a function.
So, that means you need to use the product rule
$$x \cdot \frac{\mathrm{d}y}{\mathrm{d}{x}} y + \frac{\mathrm{d}y}{\mathrm{d}{x}} x \cdot y$$
Now, what is the derivative of $y$? This is $y'$. What is the derivative of $x$? It is $1$. So, we have:
$$2(xy' + y) = 0$$
Let me know if you have further questions
$\endgroup$ $\begingroup$We have \begin{align} \frac{d}{dx}(2xy) &= 2 \left(\frac{d}{dx} (xy) \right) \\ &= 2 \left( \left(\frac{d}{dx} x \right)y + x \frac{d}{dx} y \right) \\ &= 2 (y + x y') \quad (*) \end{align} due to the rule for constants and the product rule.
This also applies for $x$ and $k$: $$ (xk)' = k + x k' $$ Only for $k' = 0$, in other words: $k$ is constant regarding $x$, this leads to $(xk)' = k$.
Because $y$ might not be constant regarding $x$, equation $(*)$ is correct.
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