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So I am taking calculus 1 online from a local college (bad idea, but the only thing that fit my schedule).

The professor used the notation $f'(x) =$ for EVERY function up until two weeks ago. All of the sudden he changed his notation to d/d$x$, d$y$/d$x$ and hasn't been able to explain it clearly enough via videos.

I am trying hard to learn how implicit differentiation works, and I've been putting in six hours per day of studying for two weeks. I just don't get it. I have read and tried the examples on MIT opencourseware, our textbook and calculus for dummies.

What is this d/d$x$ that gets placed everywhere? Where do you place it? Why use it? Same with $\text{d}y/\text{d}x$.

I've had lots of success here, so if someone could show me how to do this i would really appreciate it. This is my last hope :)

Thanks

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  • $\begingroup$ It would be helpful if you made two separate posts for your two separate questions. Otherwise things just get convoluted. $\endgroup$ – Bridgeburners Jul 31 '14 at 18:16
  • $\begingroup$ I removed the example and I just want someone to show me step by step how to use implicit differentiation. Thanks for the feedback. $\endgroup$ – Andrew Jul 31 '14 at 18:18
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To answer your first question,

$\frac d{dx}$ is an operator. So let your function be $f(x)$. The notation $\frac d{dx}f(x)$ is equivalent to saying "the derivative of $f$ with respect to $x$". Often times we will see $\frac {dy}{dx}$. This is similar. It is just "the derivative of a function of the $y$ value with respect to $x$".

For the next question,

implicit differentiation is used when it is not necessarily easier to simplify the equation first. However, we can differentiate the equation implicitly and then solve for the derivative. For an easy example:

$y^2 + x = 1$ apply $\frac d{dx}$

$\frac d{dx} (y^2 +x) = \frac d{dx} (1)$

Here the function $y$ is actually a function dependent on $x$. So we would formally write $y(x)$.

So, when we take the derivative we can simply use product rule!

$\frac d{dx}(y(x) \cdot y(x)) = y'(x)\cdot y(x) + y(x)\cdot y'(x) = 2y(x)y'(x)$

And then differentiate all other terms with respect to $x$.

$2yy' + 1 = 0$

Now it is possible to solve for $y'(x)$ which was what we were looking for! This is also the "derivative of $y(x)$ with respect to $x$" which we just learned is equivalent to $\frac d{dx} {y(x)}= \frac {dy}{dx}$!

Hopefully this cleared some things up.

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  • $\begingroup$ That is very helpful, but how do I know where to put the d/dx? Why is the d/dx before the (y^2+x) in parenthesis and before the 1? $\endgroup$ – Andrew Aug 1 '14 at 4:53
  • $\begingroup$ @Andrew If we are implicitly differentiating then we differentiate the whole equation (much like if we wanted to multiply a polynomial by 2, to keep the equation equal we should multiply both sides of the equation). The operator d/dx is just a way to symbolize a derivative. So instead of f'(x) you can write df/dx or d/dx (f(x)). Another analogy of d/dx is to say "take the derivative of what's in the parentheses ()". It takes some getting used to but, much like derivatives, you'll get it. $\endgroup$ – Eoin Aug 1 '14 at 5:21
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What is the derivative of a function? It's measure of the slope, or rate of change of a function. So what is $f'(x)$? it's the limit of the ratio of difference in height (difference in $f(x)$), to the difference in $x$, as the differences go to zero.

If you think of the $d$ in $dx$ as shorthand for "difference", then you immediately see that $df/dx$ is a very suggestive notation for $f'(x)$.

$\quad\quad\quad\quad\quad\quad\quad$enter image description here

Now, what about implicit differentiation? Even though $df/dx$ is not a real fraction, just a good notation, pertending it is a fraction leads to the law of implicit differentiation: $$\frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}$$ Or in our older notation: $$f'\left(y(x)\right) \text{(derivative acc. to $x$)}=f'(y) y'(x)$$ Now you see the strength of the $df/dx$ notation - there is no confusion about what variable the function is being derived by.

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This question is several years old, but calculus notation can be confusing to newcomers and I thought a layman's non-technical explanation might be useful for some readers.

First, consider the symbol $d \over dx$. The $d$'s in the symbol just refer to differentiation. Notice that there is an $x$ by the $d$ in the "denominator" but there is nothing beside the $d$ in the "numerator." You can read this symbol as "the derivative of ..." When you see this symbol used, there will be a function (for example, $f(x) = x^2 + 3$) either placed next to the symbol, like this: $ \frac {d}{dx} (x^2+3)$, or placed in the "numerator", like this: $ \frac {d(x^2+3)}{dx}$. Either way, this is simply read "the derivative with respect to $x$ of $x^2+3$". The same thing might be expressed as $d f(x) \over dx$, which may be read as "the derivative with respect to $x$ of $f(x)$".

Obviously, you would want to say something about the derivative with respect to $x$ of the function $f(x)$. So you might complete the mathematical statement like so:

$$\frac {d (x^2+3)}{dx} = 2x$$

This statement would be read as "the derivative with respect to $x$ of $x^2+3$ is $2x$". Or, you could say "the derivative with respect to $x$ of $f(x)$ is $2x$".

You will hear people refer to this symbol ($d \over dx$) as an operator which can be interpreted as saying that "we are taking the derivative of this function", or "the derivative of this function is ..." (whatever that function may be).

So, in response to one of Andrew's questions above (contained in a comment) "how do I know where to put the $d \over dx$?", the answer is, you put it wherever you intend to say "the derivative of ... ". For example, to express mathematically that the derivative of $x^3 - 5x$ is $3x^2-5$, you might write:

$$\frac {d}{dx}(x^3-5x) = 3x^2-5$$

You will commonly see the "prime notation" used in the same way. If the function above was defined as $y = x^2 + 3$ instead of $f(x) = x^2 + 3$, then you would commonly see the following alternative notation:

$$y^{'} = 2x$$

The same thing could be expressed as

$$\frac {d (y)}{dx} = 2x$$

In this case, the notation $y^{'}$ means "the derivative of the function $y$". Here the symbol used does not tell you the "with respect to" variable, but in this case it is safe to assume that it is the derivative with respect to $x$. Similarly, the notation $\frac {d (y)}{dx}$ can be spoken as "the derivative with respect to $x$ of the function $y$". So, whenever you want to express mathematically "the derivative of something" then you can say that by using the $d \over dx$ notation (or $d \over dy$ or $d \over d \theta$ or whatever the case may be) along with the function that will be, or has been, differentiated.

Also, referring to Eoin's answer above where Eoin showed

$$ \frac {d}{dx}(y^2+x)= \frac {d}{dx}(1)$$

Andrew asked, "why is the $d \over dx$ before the $(y^2 + x)$ and before the 1?". Eoin was explaining the process of implicit differentiation and how you must differentiate both sides of the equation. He was simply stating mathematically that "the derivative with respect to $x$ of the function $y^2+x$ is equal to the derivative of the constant value 1." At that point Eoin had not actually taken the derivative of either side of the equation. He was simply showing the step indicating that you must differentiate both sides of the equation. Therefore, his notation says that the derivative of the left side of the equation is equal to the derivative of the right side.

Now consider the symbol $dy \over dx$. As noted above (with parentheses), this symbol refers to "the derivative of $y$ with respect to $x$." It is interesting that in calculus you learn to "solve for functions" just as you have solved for variables in the past.

Here's a quiz. Without knowing what the function $f$ is, tell me its derivative with respect to time. Well, it's $df \over dt$. What is the derivative of function $p$ with respect to variable $r$? It's $dp \over dr$. What's the derivative of $\sigma$ with respect to $\omega$? It's $d \sigma \over d \omega$. You don't know what the functions actually are, but you can express them with a symbol and then, when given more information, solve for them.

For example, given the rate at which the radius of a circle is changing (i.e., the derivative of radius with respect to time, $dr \over dt$), you can calculate the rate at which the area is changing (i.e., the derivative of area with respect to time, $dA \over dt$). Similarly, given the rate at which the area is changing, you can calculate the rate at which the radius is changing. Just find a function relating the two things (area and radius of a circle in this case) and then differentiate (in this case, with respect to time):

$$A = \pi r^2$$ $$\frac {d}{dt}(A) = \frac {d}{dt}(\pi r^2)$$ $$\frac {dA}{dt} = 2 \pi r \frac {dr}{dt}$$

(The first line gives the equation to be differentiated. The second line says that I am going to differentiate both sides. The third line actually performs the differentiation.) In this case, you probably don't yet know the rate of change of the area (i.e., $dA \over dt$), or the rate of change of radius (i.e., $dr \over dt$) but that does not stop you from representing those values with a symbol. You also could simplify the notation somewhat if you chose to do so - something like this, perhaps:

$$A = \pi r^2$$ $$A^{'} = \frac {d}{dt}(\pi r^2)$$ $$A^{'} = 2 \pi r r^{'}$$

Both equations say exactly the same thing, namely, the derivative of the area of the circle with respect to time (i.e., its rate of change) is 2 times $\pi$ times the radius times the derivative of the radius with respect to time (i.e., its rate of change). Given $A^{'}$ or $r^{'}$ you can solve for the other.

I hope that helps clarify some of the notation for students new to calculus.

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If you give an increment to $x$, let $\Delta x$, the corresponding increment of $f(x)$ is $\Delta f(x)$ and if you let $\Delta x$ tend to zero, then $\dfrac{\Delta f(x)}{\Delta x}$ tends to $f'(x)$ by definition (if that limit exists).

$\Delta f(x)$ is the true variation of $f(x)$ corresponding to the variation of $x$ and when $f'(x)$ exists, $\Delta f(x)$ has a part proportional to $\Delta x$ and a remainder:

$$\Delta f(x)=f'(x)\Delta x+R(x,\Delta x).$$

Usage has enforced the notation

$$df(x)=f'(x)\,dx$$ to denote the linear part, where $dx$ needn't tend to zero, and for this reason,

$$\frac{df(x)}{dx}=f'(x)$$ holds whatever $dx$.


Some will claim that $dx$ is an "infinitesimal quantity", but this is not true with the modern definition of the differential.

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