# Calculus: Meaning of the differentiate sign $\frac{d}{dx}$, Why is $\frac{d}{dx}(\sin y)$ applied with chain rule but $\frac{d}{dx}(\sin x) =\cos(x)$?

1. I'm getting confused with the different signs.

I understand that $$\frac{dy}{dx}$$ reads "$$y$$ in respect to the derivative of $$x$$" & $$\frac{d}{dx}$$ is to differentiate a certain equation. But I don't think I really understand what it means when another function is mixed in apart of $$x$$.

A more concrete example of what I don't understand is:

• Why is $$\frac{d}{dx}(\sin y)$$ applied with chain rule but $$\frac{d}{dx}(\sin x) =\cos(x)$$ applies the derivative-of-sine "rule"
• $$\frac{d}{dx}(y^2)$$ applies the chain rule, but $$\frac{d}{dx}(x^2)$$ applies the power rule
2. sub-question: There's so many rules in derivatives of calculus alone, do I have to remember them all, will formula usually be provided? (ie. power rule, chain rule I get. But derivatives of $$\sin(x)$$, $$\cos(x)$$, $$e^x$$, $$\ln(x)$$, $$\tan(x)$$, $$\cot(x)$$, $$\sec(x)$$, $$\csc(x)$$, $$\arcsin(x)$$..?)

• If I have to remember them, is there a few main ones to remember then derive them after.
• think of $y$ as a function of $x$ – J. W. Tanner Oct 18 '20 at 3:30
• I question whether relying on your memory is a good idea. I think studying math analogizes to driving a car on the freeway. If you learn how to drive so that it is second nature, your chances of not crashing the car improve. In math, the idea would be to attack the problems in your textbook, and then (if needed) go 1-on-1 with a teach or tutor. This approach will stretch your intuition and lead to understanding the topic, which will then lead to learning. – user2661923 Oct 18 '20 at 3:51

I have been in your shoes at one point, and I think that your problem is that you are thinking of derivative as finding slopes. I instead I present a new abstract interpretation which will possibly dispel all doubts.

If we have an equation of the sort:

$$x+3 = 2$$

Then we can rearrange it to get:

$$x=-1$$

Similarly, if have an equation with two variables (implicit curve) then we apply the $$\frac{d}{dx}$$ operator on both sides to find to relate the rate of change of variables. For example, consider the equation of a circle:

$$x^2 + y^2 =1$$

If we apply $$\frac{d}{dx}$$

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

Now here we say that to keep on the circle, when we change our $$x$$ the $$y$$ must change as a function of it. Considering that and simplifying,

$$\frac{d}{dx} x^2 + \frac{d}{dx} y^2 =0$$

Or,

$$2x + 2y y' = 0$$

Or,

$$y' = -\frac{x}{y}$$

WIth that in mind,

$$\frac{d}{dx} \sin y = \frac{d}{dx} ( \sin x) |_y \frac{dy}{dx}$$

Is due to the fact that we are saying that $$y$$ is a function of $$x$$. The real idea behind applying the chain rule when the $$y$$ is inside because we want to say that $$y$$ is dependent on $$x$$ but if we said both variables were not correlated at all i.e: you could freely change $$x$$ and $$y$$ independent of each other then the derivative would be zero.

Remembering identities

First of all I suggest that you try to derive all the identities by yourself from scratch. However, there is an easy way to derive the inverse identities by a clever application of the chain rule. I have written about it here.

For learning more about $$d$$ as an operator, see here

• I thought derivatives are slopes? I'm not sure if I get the alternative you're suggesting. I follow the algebra all the way till $\frac{d}{dx}\sin y = \frac{d}{dx}\sin x |_y \frac{dy}{dx}$ . in particular, $|_y$ before $\frac{dy}{dx}$ I don't think I'm in at the point in my learning yet. If I were to guess you're essentially stating the chain rule here, but replacing the sin y with sin x on the right side of the equation? Why? From my rote learning $\frac{d}{dx}\sin y = \frac{d}{dx}\sin y * \frac{d}{dx} (y)$. /// Thanks for the video link btw. Are formulas usually provided in exams? – nvs0000 Oct 18 '20 at 11:50
• Derivatives have many other interpetation. THe slope is the easiest to explain. The $|_y$ is trying to denote the idea that after we take derivative we evaluate the function at $y$. In my school exams, the formula weren't provided. – Buraian Oct 18 '20 at 11:55
• $$\frac{d}{dx} \sin x = \cos x$$ Now evaluate that at x = y – Buraian Oct 18 '20 at 11:58

To address question 1. it might be useful to think that the chain rule always applies. Consider the derivative of $$\sin(y)$$ with respect to $$x$$, we find $$\frac{\mathbf{d}}{\mathbf{d}x}(\sin(y)) = \sin'(y)\frac{\mathbf{d}y}{\mathbf{d}x} = \cos(y)\frac{\mathbf{d}y}{\mathbf{d}x}$$ But now suppose $$y=x$$, then $$\frac{\mathbf{d}y}{\mathbf{d}x} = \frac{\mathbf{d}x}{\mathbf{d}x} = 1$$, so

$$\frac{\mathbf{d}}{\mathbf{d}x}(\sin(y)) =\cos(y)\frac{\mathbf{d}y}{\mathbf{d}x} = \cos(x)\frac{\mathbf{d}x}{\mathbf{d}x}= \cos(x)$$ Hope this helps a little. As for question 2, practice makes perfect and with time you will find it manageable to remember all the necessary derivatives and derivative rules.

• @olif9837 I suggest that when you are examining this answer, you pretend that (for example) $y = x^2$, and then compare $\frac{d}{dx} \sin (y)$ with $\frac{d}{dx} \sin (x)$. – user2661923 Oct 18 '20 at 3:55

The sign $$\frac{d}{dx}$$ is "derivative with respect to $$x$$", in the way you are using it $$y$$ represent a function with respect to $$x$$ (can be any function, $$e^x$$, $$ln(x)$$, $$x^n$$, etc.) For example if $$y=e^x$$, $$\frac{d}{dx}(\sin y) = \frac{d}{dx} (\sin (e^x))$$, for this you need to use the chain rule, in this particular case: $$\frac{d}{dx}(\sin y)=\cos(y)\cdot y^{\prime}= \cos(e^x) \cdot e^x$$

For the other question, no, there are many rules, but is "easy" to deduce ones from others, for example if you know that $$\tan = \frac{\sin}{\cos}$$ you don't need to memorize the rule for $$\tan$$, you probably need the quotient rule, the product rule and the chain rule. In this example you know that $$\frac{d}{dx}\frac{f(x)}{g(x)}= \frac{f^{\prime}(x)\cdot g(x)-g^{\prime}(x)\cdot f(x)}{g(x)^2}$$ With this you can deduce that $$\frac{d}{dx} \tan x = \frac{d}{dx} \frac{\sin x}{\cos x} = \frac{\cos x \cdot\cos x - (-\sin x)\sin x}{\cos^2 x }=\frac{1}{\cos^2 x}=\sec ^2 x,$$ and this is exactly the "rule" from $$\tan x$$. So it is not necessary to memorize every single rule, but you need some of them.

• Applying the quotient rule to $\tan$ results in $1/ \cos^2$. – Jens Schwaiger Oct 18 '20 at 4:23

The way I remember it is to separate the actions of "differentiate" and "divide by $$dx$$". I use $$d()$$ as the differential operator, and $$\frac{d}{dx}\left(\right)$$ combines the differential operator and "divide by $$dx$$".

So, for instance, $$d(x^2) = 2x\,dx$$. However, for $$\frac{d}{dx}(x^2)$$, you divide by $$dx$$ afterwards, giving $$2x$$.

Now, let's look at a similar one. If we have $$d(y^2)$$ we have $$2y\,dy$$. However, for $$\frac{d}{dx}(y^2)$$ you divide by $$dx$$ afterwards, giving $$2y\,\frac{dy}{dx}$$.

The way I always do it is to differentiate first, then just solve for whichever derivative I'm looking for (i.e., solve for the ratio of differentials that I'm looking for).