When does $\frac{dy}{dx}= \frac{dy}{du} \frac{du}{dx}$?

Question

Lets say we have $y = \frac{1}{2}x - u + u^2$.

Obviously $\frac{dy}{dx} = \frac{1}{2}$ but I've seen many problems that rely on $\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx} = \frac{dy}{du} \div \frac{dx}{du}$

But, from what I can tell, this won't work for many equations like the above one.

$\frac{dy}{du} = -1 + 2u$

& if you rearrange $y = \frac{1}{2}x - u + u^2$ for x, you get $x = 2y +2u - 2u^2$ with $\frac{dx}{du} = 2 - 4u$

$$\begin{align} \frac{dy}{du} \div \frac{dx}{du}&= \frac{2u - 1}{2 - 4u}\\ &= \frac{2u - 1}{-2(2u -1)}= -\frac{1}{2} \end{align}$$

So I guess I'm a little confused as to when you can use $\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$ & when you can't?

Thanks for any help

Answer 1

The problem here is that you seem only to "understand" these things mechanically. For example, consider the equation $$y=\frac12x-u-u^2$$ which you gave. You just started working on it without considering what you meant by it, hence the nonsensical results.

Now before you can do calculus on this equation, you must understand that it has to somehow define a function; that has a very specific meaning, so you may want to check it out. Coming back to your example, it contains three variables, so if it wants to be used to define a function, then it must be a function of at least two variables (assuming, of course, that all variables are independent; otherwise we can reduce it to a function of one variable). So if we take $y(x,u)$ to define some function $f$, then you can't handle things as cavalierly as you did. I don't know whether you're familiar with several variable calculus, so I'll just pick another function to apply the chain rule to. I'll use your equation again, to show you that symbols by themselves mean nothing or all things until we define what they mean to us. So consider the function $g$ given as $$g(x)=\frac12x-u-u^2,$$ where $u$ is some constant. Then you can say that the instantaneous rate of change of $g$ with $x$ at any point in its domain is given by $$\frac{dy}{dx}=1/2.$$ In this case, it makes no sense to use the chain rule since $x$ was assumed to be independent, so that it does not depend on some other variable.

Now consider yet again your equation, and now (with a little change, as you'll see) let it define a function $$h(z)=\frac12x+z,$$ where $z=u^2-u$. Note that now, $z$ is the variable while $x$ is a constant value. Then it is easy to see that $z$ by itself also defines some function of $u$, so that we may write $h$ as $h(z(u))$. This is what the chain rule is for. Now let us differentiate $h$ wrt $z$, to give $$\frac{dh}{dz}=1,$$ and similarly, $$\frac{dz}{du}=2u-1.$$ Now, what if we wanted to see how $h$ changes with $u$ instead? Of course we could just substitute for $z$ in $h(z)$ and work things out, but in many cases it is undesirable as it involves additional computational steps, which the chain rule helps us bypass.

The chain rule is that for any two differentiable functions $\phi(\gamma)$ and $\gamma(x)$, where $x$ is an independent variable, $$\frac{d\phi}{dx}=\frac{d\phi}{d\gamma} \frac{d\gamma}{dx}.$$ This is true in general because we have a proof, which is simple enough to follow. Now to answer your question, the chain rule is true because we can write, for a change $\Delta x$ in the independent variable $x$, $$\frac{\Delta\phi}{\Delta x}=\frac{\Delta\phi}{\Delta \gamma} \frac{\Delta\gamma}{\Delta x},$$ so that when we let $\Delta x\to 0$, we obtain the chain rule $$\frac{d\phi}{dx}=\frac{d\phi}{d\gamma} \frac{d\gamma}{dx}.$$

Now let us go back to the example I gave above where we wanted to know $dh/du$. By the chain rule, $$\frac{dh}{du}=\frac{dh}{dz} \frac{dz}{du},$$ so that we have (from our previous computation) $$\frac{dh}{du}= 1\cdot (2u-1)=2u-1,$$ which gives the same result if we'd substituted first for $z$ in $h(z)$.

Answer 2

The chain rule $$\frac{dy}{dx}= \frac{dy}{du} \frac{du}{dx}$$ is used when y is a function of u and u is a function of x.

That is if y is a composite function, $ y=y(u(x))$

For example in the statement of $$ y=u^2 + 3u, u=x^3-5x $$ we get $$\frac{dy}{dx}=(2u+3)(3x^2-5)=(2x^3-10x+3)(3x^2-5)$$

Answer 3

Whenever all three three quantities $\frac{\mathrm{d}y}{\mathrm{d}x}$, $\frac{\mathrm{d}y}{\mathrm{d}u}$, and $\frac{\mathrm{d}u}{\mathrm{d}x}$ make sense, you always have an equation

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x} $$

Furthermore, whenever both quantities $\frac{\mathrm{d}u}{\mathrm{d}x}$ and $\frac{\mathrm{d}x}{\mathrm{d}u}$ make sense, you have an identity

$$\frac{\mathrm{d}u}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}u} = 1$$

The quantity $\frac{\mathrm{d}y}{\mathrm{d}x}$, for example, only makes sense when $x$ and $y$ are related. For example, this is true when there is a function $f$ such that $y = f(x)$. (this paragraph should be qualified by including smoothness conditions)

In your given problem, $\frac{\mathrm{d}y}{\mathrm{d}x}$ only makes sense if $u$ is also related to $x$. If $u$ is not constant, then you've you've incorrectly computed the derivative: it should be

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2} - \frac{\mathrm{d}u}{\mathrm{d}x} + 2 u \frac{\mathrm{d}u}{\mathrm{d}x} $$

Note, incidentally, that the differentials make sense no matter how the variables are or are not related: given your equation, it's always true (again assuming smoothness conditions) that

$$ \mathrm{d}y = \frac{1}{2} {\mathrm{d}x} + \left( 2 u - 1 \right) \mathrm{d}u $$

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Note that notation such as $\frac{\partial y}{\partial x}$ is something else entirely; great care must be used when using this notation, and it is very fragile with respect to "creative" manipulation. For example, you should rarely expect any reasonable relationship between what you mean by $\frac{\partial y}{\partial x}$ and what you mean by $\frac{\partial x}{\partial y}$. Detail is beyond the scope of this post.