How do you show that the derivative of cosx = -sinx?

Question

Show that the derivative of cosx = -sinx given that

sinx = (e^(ix) - e^(-ix))/(2i) and cosx = (e^(ix) + e^(-ix))/2 .

Answer 1

The first answer is correct. Another method is to show cos(x) as a series which is

cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6! + x⁸/8! ..............

d(cos(x))/dx = -x + x³/3! - x⁵/5! + x⁷/7! .............. which is the series for -sin(x)

Answer 2

Starting with the given definition for cos(x), we write

cos(x) = (e^ix + e^-ix)/2

Differentiation with respect to x yields

d(cos(x))/dx = d/dx ((e^ix + e^-ix)/2)

d(cos(x))/dx = (d(e^ix)/dx + d(e^-ix)/dx)/2

We know that the derivative of e^x with respect to x is just e^x itself. So, also using chain rule, we can write

d(cos(x))/dx = (e^ixd(ix)/dx + e^-ixd(-ix)/dx)/2

d(cos(x))/dx = (ie^ix - ie^-ix)/2

d(cos(x))/dx = i(e^ix - e^-ix)/2

We know that i³ = -i which means i = -1/i. So,

d(cos(x))/dx = -(e^ix - e^-ix)/(2i)

It is also given that (e^ix - e^-ix)/(2i) = sin(x). Therefore,

d(cos(x))/dx = -sin(x)