How do you differentiate $cos 2 x$ $cos 2x$ ?

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Answer 1 · 2017-01-03T01:13:00

We let $y = cos u$ $y = cosu$ and $u = 2 x$ $u = 2x$ . Then $\frac{d y}{d u} = - sin u$ $dy/(du) = -sinu$ and $\frac{d u}{d x} = 2$ $(du)/dx= 2$ .

$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ $dy/dx = dy/(du) xx (du)/dx$

$\frac{d y}{d x} = - sin u \times 2$ $dy/dx = -sinu xx 2$

$\frac{d y}{d x} = - 2 sin u$ $dy/dx = -2sinu$

Since $u = 2 x$ $u = 2x$ :

$\frac{d y}{d x} = - 2 sin 2 x$ $dy/dx = -2sin2x$

Hopefully this helps!

How do you differentiate cos2xcos 2x?