Need explanation for intuition behind rewriting $dy$ in terms of $dx$

Question

Khan Academy provides the following article and intuition behind rewriting $dy$ in terms of $dx$. I do not understand the intuition... If we move $dx$ to the right, the amount we need to step along the $y$ axis would be $sin(x + dx) - sin(x)$ (which I am not sure how to rewrite, I already forgot trig operations..., but I don't see how we get $cos(x) dx$ even intuitively). How do we get $cos(x) dx$ ??

Answer 1

Just $dy=y'dx$.

For example, $$d(\cos{x})=(\cos{x})'dx=-\sin{x}dx$$ or $$d(\sin{x})=(\sin{x})'dx=\cos{x}dx.$$

I think it means the following. $$\sin(x+\Delta x)-\sin{x}=2\cos\left(x+\frac{\Delta x}{2}\right)\sin\frac{\Delta x}{2}\sim\cos{x}\Delta x$$ because $2\sin\frac{\Delta x}{2}\sim\Delta x.$

Answer 2

Precisely speaking, the change in $y$ is $\sin(x+\Delta x)-\sin(x)$. Using trig identities you get $\sin(x+\Delta x)=\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x)$, so the difference is $\sin(x)(\cos(\Delta x)-1)+\cos(x)\sin(\Delta x)$. For small $\Delta x$ you approximate $\sin(\Delta x) \approx \Delta x$ and $\cos(\Delta x)-1 \approx 0$, both of which can be understood using just trigonometry. (Note that the latter means that $\cos(\Delta x)-1$ is not just going to zero but is going to zero "faster than $\Delta x$", in the sense that $\lim_{\Delta x \to 0} \frac{\cos(\Delta x)-1}{\Delta x}=0$.) This gives the desired result.

The intuition is that the derivative is the slope of the tangent line, and going from $(x,y)$ to $(x+dx,x+y'(x)dx)$ is exactly following the tangent line a distance $dx$ along the $x$ axis, which very nearly follows the curve itself if $dx$ is very small.

Answer 3

$$\begin{align} y &= \sin x \\ \frac{dy}{dx} &= \frac{d}{dx}(\sin x) \\ dy &= d(\sin x) \end{align}$$

Instead of eliminating the $dx$ in either side first, you could do this:

$$\begin{align} \frac{dy}{dx} &= \frac{d}{dx}(\sin x) \\ &= \cos x \\ dy &= \cos(x) \, dx \\ \end{align}$$

Moving around the $dx$ is just notation, though some people say $f(x)\,dx$ is meaningless by itself because it’s tinker than any real amount (other than zero). In reality, $dy=f(x)\,dx$ is a synonym to saying $dy/dx=f(x)$, nothing more. Actually, that technique is something you use in $u$-substitution all the time.

Does that help?

Answer 4

If you have a smooth function $ y=f(x)$, then we use two different words to measure the rate of change and the linearized change at a given point $x$.

The first word is the derivative which is the instantaneous rate of change and we denote it by either $y'$ or by $f'(x)$ or by $ dy/dx $ or by $df/dx$.

The second word is differential, $dy $ which is the linearized change at the point $x$.

That is a linear approximation to the actual change which is $\Delta f =f(x+dx)-f(x)$

In order to find the linear approximation, we multiply the instantaneous rate of change at $x$ by the $x$ increment $dx$ that is $df= f'(x) dx$, or $dy = y' dx.$

Thus the differential is the derivative times the $x$ increment.

For $ f(x) = sin(x)$, the derivative is $f'(x) = cos(x)$ and the differential is $df=cos(x)dx.$

Answer 5

As indicated in the diagram, you are at some point whose abscissa is $x$, therefore the point itself is $(x,\sin x)$. You move some distance $dx$ to the left(you can also have $dx$ moving to the right, but I chose left), but want to stay on the graph. You know that then, you land up on the point $(x+dx,\sin(x + dx))$. The question is the following : what is $?$

What Mr. Khan is trying to say is exactly this : he is trying to tell you that $?$ is the displacement $dx$, times some function of $x$, which is exactly the derivative of $\sin x$.

But why does the derivative come up in the first place?

Abstractly, the derivative of a function $f$ at a point $x$ is defined as $f'(x )=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$. Put $f=\sin $ and for your convenience, let's rename $h$ as $dx$ to get $f'(x) = \lim_{dx \to 0} \frac{\sin(x+dx) - \sin x}{dx}$.

Now, provided $dx$ is very small, $f'(x)dx \approx \sin(x+dx) - \sin x$, and indeed, from the diagram it is clear that $?$ is the difference of the $y$-coordinates, which is just $\sin(x+dx) - \sin x$.

Therefore, given any function $f$ differentiable at a point $x$, it is true that $f'(x)dx \approx f(x+dx) -f(x)$. Therefore, if displaced from $x$ by $dx$ units, the number of units up or down you must move is $f'(x)dx$.

The calculation for the derivative of $\sin x$ is already done by other users, hence I will not attempt it. My job was to bring the idea of why the intuition works in defining the derivative. I hope the diagram was helpful.