How do you differentiate $2 {cos}^{2} (x)$ $2cos^2(x)$ ?

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Answer 1 · 2017-01-19T08:10:51

$\frac{d y}{d x} = - 4 cos x sin x$ $(dy)/(dx) =-4cosxsinx$

To differentiate $y = 2 {cos}^{2} x$ $y=2cos^2x$ we need the chain rule

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

let $u = cos x \Rightarrow y = 2 u^{2}$ $u=cosx=>y=2u^2$

$\frac{d u}{d x} = - sin x$ $(du)/(dx)=-sinx$

$\frac{d y}{d u} = 4 u$ $(dy)/(du)=4u$

$:.(dy)/(dx)=4uxx(-sinx)$

substitute back for $u$

$(dy)/(dx) =-4cosxsinx$

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