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

Question

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.

Answer 1

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.

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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

Answer 2

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.

Answer 3

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.

Answer 4

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).