Question

Find the derivative of y = cos(ax - b) from the first princip?​

Answer 1

Recall that

sin(x ± y) = sin(x) cos(y) ± cos(x) sin(y)

cos(x ± y) = cos(x) cos(y) ∓ sin(x) sin(y)

Then

cos(ax - b) = cos(ax) cos(b) + sin(ax) sin(b)

When you differentiate y with respect to x, you only need to focus on the the cos(ax) and sin(ax) terms.

We have

$y = \cos(ax-b)$

$\displaystyle\frac{\mathrm dy}{\mathrm dx} = \cos(b)\lim_{h\to0}\frac{\cos(a(x+h))-\cos(ax)}h + \sin(b)\lim_{h\to0}\frac{\sin(a(x+h))-\sin(ax)}h$

$\displaystyle\frac{\mathrm dy}{\mathrm dx} = \cos(b)\lim_{h\to0}\frac{\cos(ax+ah)-\cos(ax)}h + \sin(b)\lim_{h\to0}\frac{\sin(ax+ah)-\sin(ax)}h$

$\displaystyle\frac{\mathrm dy}{\mathrm dx} = \cos(b)\lim_{h\to0}\frac{\cos(ax)\cos(ah)-\sin(ax)\sin(ah)-\cos(ax)}h \\+ \sin(b)\lim_{h\to0}\frac{\sin(ax)\cos(ah)+\cos(ax)\sin(ah)-\sin(ax)}h$

$\displaystyle\frac{\mathrm dy}{\mathrm dx} = \cos(b)\lim_{h\to0}\frac{\cos(ax)(\cos(ah)-1)-\sin(ax)\sin(ah)}h \\\\+ \sin(b)\lim_{h\to0}\frac{\sin(ax)(\cos(ah)-1)+\cos(ax)\sin(ah)}h$

$\displaystyle\frac{\mathrm dy}{\mathrm dx} = \cos(ax)\cos(b)\lim_{h\to0}\frac{\cos(ah)-1}h\\\\-\sin(ax)\cos(b)\lim_{h\to0}\frac{\sin(ah)}h\\\\ + \sin(ax)\sin(b)\lim_{h\to0}\frac{\cos(ah)-1}h\\\\+\cos(ax)\sin(b)\lim_{h\to0}\frac{\sin(ah)}h$

Now, recall these useful known limits: for c ≠ 0,

$\displaystyle\lim_{x\to0}\frac{\sin(cx)}{cx}=1\text{ and }\lim_{x\to0}\frac{1-\cos(cx)}{cx}=0$

Then the limits involving cosine vanish, and the derivative simplifies to

$\displaystyle\frac{\mathrm dy}{\mathrm dx} = -\sin(ax)\cos(b)\lim_{h\to0}\frac{\sin(ah)}h +\cos(ax)\sin(b)\lim_{h\to0}\frac{\sin(ah)}h$

For the remaining limits, introduce a factor of a in the denominators:

$\displaystyle\frac{\mathrm dy}{\mathrm dx} = -a\sin(ax)\cos(b)\lim_{h\to0}\frac{\sin(ah)}{ah} + a\cos(ax)\sin(b)\lim_{h\to0}\frac{\sin(ah)}{ah}$

$\displaystyle\frac{\mathrm dy}{\mathrm dx} = -a\sin(ax)\cos(b) + a\cos(ax)\sin(b)$

and using the first identity listed above, we can write this as

$\displaystyle\frac{\mathrm dy}{\mathrm dx} = -a(\sin(ax)\cos(b) - \cos(ax)\sin(b)) = \boxed{-a\sin(ax-b)}$