How do you differentiate ${cos (1 - 2 x)}^{2}$ $cos(1-2x)^2$ ?

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Answer 1 · 2018-05-08T11:03:58

$\frac{d y}{d x} = 4 cos (1 - 2 x) sin (1 - 2 x)$ $dy/dx=4cos(1-2x)sin(1-2x)$

First, let $cos (1 - 2 x) = u$ $cos(1-2x)=u$

So, $y = u^{2}$ $y=u^2$

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

$\frac{d y}{d u} = 2 u$ $(dy)/(du)=2u$

$\frac{d u}{d x} = \frac{d}{d x} [cos (1 - 2 x)] = \frac{d}{d x} [cos (v)]$ $(du)/(dx)=d/dx[cos(1-2x)]=d/dx[cos(v)]$

$\frac{d u}{d x} = \frac{d u}{d v} \cdot \frac{d v}{d x}$ $(du)/(dx)=(du)/(dv)*(dv)/(dx)$

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d x}$ $dy/dx=(dy)/(du)* (du)/(dv) *(dv)/(dx)$

$\frac{d u}{d v} = - sin (v)$ $(du)/(dv)=-sin(v)$
$\frac{d v}{d x} = - 2$ $(dv)/(dx)=-2$

$\frac{d y}{d x} = 2 u \cdot - sin (v) \cdot - 2$ $dy/dx=2u*-sin(v)*-2$

$\frac{d y}{d x} = 4 u sin (v)$ $dy/dx=4usin(v)$

$\frac{d y}{d x} = 4 cos (1 - 2 x) sin (1 - 2 x)$ $dy/dx=4cos(1-2x)sin(1-2x)$

How do you differentiate cos(1−2x)2cos(1-2x)^2?