Proving the derivative $\sin'(kx) = k\cos(kx)$ using the fundamental limit of $\frac{\sin(\theta)}{\theta} = 1$ as $\theta\to 0$

Question

I started off by using the fundamental definition of a derivative:

• Let $f(x) = \sin(kx)$
• $f'(x) = \lim_{h\to 0} \frac{\sin(k(x+h)) - \sin(kx)}{k(x+h)-kx}$

Which by the coumpound angle formula:

• $f'(x) = k^{-1}(\lim_{h\to 0} \frac{\sin(kx)\cos(kh) + \cos(kx)\sin(kh) - \sin(kx)}{h})$
• $f'(x) = k^{-1}(\lim_{h\to 0} \frac{\sin(kx)\cos(kh)- \sin(kx)}{h} + \lim_{h\to 0} \frac{\cos(kx)\sin(kh)}{h})$
• $f'(x) = k^{-1}( \sin(kx)(0) + \cos(kx)(1))$
• $f'(x) = k^{-1}\cos(kx)$

Which is not correct. The $k^{-1}$ should be $k^1$.

Answer 1

Let $f(x) = \sin(kx)$. Then its derivative is given by

\begin{align*} f'(a) & = \lim_{x\to a}\frac{f(x) - f(a)}{x - a}\\\\ & = \lim_{x\to a}\frac{\sin(kx) - \sin(ka)}{x - a}\\\\ & = \lim_{x\to a}\frac{2\sin\left(\dfrac{k(x-a)}{2}\right)\cos\left(\dfrac{k(x + a)}{2}\right)}{x - a}\\\\ & = \lim_{x\to a}\frac{\sin\left(\dfrac{k(x - a)}{2}\right)}{k\left(\dfrac{x-a}{2}\right)}\times\lim_{x\to a}k\cos\left(\dfrac{k(x + a)}{2}\right)\\\\ & = k\cos\left(ka\right) \end{align*}

where we have applied the proposed fundamental limit and the fact that $\cos$ is continuous.

Hopefully this helps !