Question

Find the derivative by first principle:
cos 5x

Answer 1

-5sinx is right answer .................

Answer 2

Answer:

$\dfrac{d(cos5x)}{dx}= -5sin5x$

Step-by-step explanation:

Since we know that according to first principal $f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$.............(1)

we have

$f(x) = cos5x \\f(x+h) = cos(5x +5h)$

So, using (1) we get,

$f'(x) = \lim_{h\to0} \frac{cos(5x+5h)-cos5x}{h}$......................(2)

we know that

$cosA-cosB = 2sin(\frac{B-A}{2})sin(\frac{A+B}{2})$.................(3)

Using (2) and (3) we get,

$f'(x) = \lim_{h\to0} \dfrac{2sin(\frac{-5h}{2})sin(\frac{10x+5h}{2})}{h}$

So,$f'(x) = \lim_{h\to0} \dfrac{\dfrac{-5}{2} 2sin(\frac{-5h}{2})sin(\frac{10x+5h}{2})}{\frac{-5}{2} h}$

Apply limit we get,

$f'(x) = {-5}sin(\frac{10x}{2})} = -5sin(5x)$  ( $\lim_{h \to 0} \frac{sinh}{h} = 1$)

$\dfrac{d(cos5x)}{dx}= -5sin5x$