How to find the derivative of $cos 5x$ using first principle

Question

So my textbook solution uses some discrete method to arrive at -5sinx 5x. But I want to know a simpler technique to get the solution. Can anyone post?

Answer 1

By the definition of a derivative, you have

$$f’(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$

So using $f(x) = \cos(5x)$, you get

$$\lim_{h \to 0}\frac{\cos(5x+h)-\cos(5x)}{h}$$

To get you started, you need to know:

• $$\cos(\alpha+\beta) = \cos (\alpha)\cos (\beta)-\sin (\alpha)\sin (\beta)$$
• $$\color{blue}{\lim_{x \to 0} \frac{\sin(x)}{x} = 1}$$
• $$\lim_{x \to 0}\frac{\cos(x)-1}{x} = 0$$

For the limit highlighted in blue, you need a simple manipulation since you get a limit in the form $\lim_\limits{x \to 0}\frac{\sin(nx)}{x}$. Can you take it on from here?

Answer 2

$$\lim_{h\rightarrow0}\dfrac{f(x+h)-f(x)}{h}$$ $$\implies \lim_{h \to 0}\frac{\cos5(x+h)-\cos5(x)}{h}$$ $$\implies \lim_{h \to 0} \frac{\left(1-\frac{1}{2}5^2(x+h)^2+\frac{1}{24}5^4(x+h)^4-.... \right)-\left((1-\frac{1}{2}5^2(x)^2+\frac{1}{24}5^4(x)^4-.... \right)}{h}$$

$$\implies \lim_{h \to 0}\frac{-\frac{1}{2}5^2(2xh+h^2)+\frac{1}{24}5^4(4x^3h+6x^2h^2+4xh^3+h^4)...-}{h}$$ $$\implies \lim_{h \to 0} \frac{h\left[-\frac{1}{2}5^2(2x+h)+\frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)...- \right]}{h}$$ $$\implies \lim_{h \to 0}\left[-\frac{1}{2}5^2(2x+h)+\frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)-... \right]$$ $$\implies -5\left[\frac{1}{2}5(2x)-\frac{1}{24}5^3(4x^3)+... \right] $$ $$\implies -5\left[(5x)-\frac{1}{6}(5x)^3+... \right] $$ $$\implies -5\sin 5x$$