Why does the derivative of $\sin(kx) = k\cos(kx)$, practically

Question

I was in a lecture in college and my lecturer mentioned the practical reason why the derivative of $ \sin(kx) = k\cos(kx) $. Although I didn't quite understand the reason totally.

Would anyone be able to explain why this derivative is so in practical terms and or logical terms, not with a mathematical proof (i.e. using the chain rule)? I was wondering if you could think of it in terms of the graphs of each of the functions?

I don't really know the reason, therefore any insight/help will be greatly appreciated.

Thank you :-)

Answer 1

Consider the graph of a function $f(x)$. If we multiply the argument of the function (the "$x$", in this case), we squash the function horizontally. For example, if we take $f(2x)$, it's as if we smooshed together the graph into half the horizontal space. You might want to test this in a graphical calculator such as desmos.com to test this for yourself.

If we smoosh it together by a factor of $k$, the slope will increase by a factor $k$ as well.

So, since $\sin'(x) = \cos(x)$, we have that the slope of $\sin(kx)$ must be $k$ times $\cos(kx)$.

Answer 2

Start with the graph $y = \sin x$. How is that related to the graph of $y=\sin(kx)$? In the graph of $y = \sin(kx)$, vertical dimensions remain the same, but horizontally everything is shrunk by a factor $1/k$. That means all the slopes are multiplied by a factor of $k$.

Here are the graphs of $\sin x$ and $\sin(2x)$.

Answer 3

Informally, when we say that the 'derivative of sine is cosine', what we mean is that the following quotient is equal to $\cos x$: $$ \frac{\sin(x+dx)-\sin(x)}{dx} $$ where $dx$ is an infinitely small non-zero number. In other words, the line joining the two points $(x,\sin x)$ and $(x+dx,\sin(x+dx))$ has a slope of $\cos x$. Symbolically, if $f(x)=\sin x$, then $f'(x)=\cos x$. Now consider the graph of $g(x):=f(kx)$. The $y$-coordinates remain the same as before, but the $x$-coordinates are scaled by a factor of $1/k$. This means that the point $(x,\sin x)$ is mapped to $(x/k,\sin x)$, since $g(x/k)=f(x)=\sin(x)$. Similarly, the point $(x+dx,\sin(x+dx))$ is mapped to $(\frac{x+dx}{k},\sin(x+dx))$. Hence, \begin{align} g'(x/k) &= \frac{\sin(x+dx)-\sin(x)}{\frac{x+dx}{k}-\frac{x}{k}} \\[4pt] &= k\cos x \, . \end{align} And if $g'(x/k)=k\cos x$, then $g'(x)=k\cos kx \; \blacksquare$

Answer 4

Not exactly an answer, but perhaps some intuition can be derived from this...

Consider the point on the plane $p_k(t) = (\cos kt, \sin kt)$. It describes a circle with 'speed' $k$ (that is, as $|k|$ increases, the point moves around the circle faster).

If you increase $|k|$ (the sign determines the direction around the circle) you can imagine that the velocity increases. It is not to hard to believe that the velocity is proportional to $k$ (I am allowing for sign of $k$ here, the point could move in either direction).

If you accept this, we just need to determine the direction.

At any point, the velocity of the point is tangential to the circle at $p_k(t)$. That is, $v \bot p_k(t)$.

Note that $(-\sin kt, \cos kt)$ is a unit vector perpendicular to $p_k(t)$.

Combining the above, we guess that $p'_k(t) = (-k\sin kt, k\cos kt)$ from which we get that the derivative of $t \mapsto \sin kt$ (the second component of $p_k(t))$ is $t \mapsto k \cos kt$.