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

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Answer 1 · 2018-05-08T11:03:58

$dy/dx=4cos(1-2x)sin(1-2x)$

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

So, $y=u^2$

$dy/dx=(dy)/(du)*(du)/(dx)$

$(dy)/(du)=2u$

$(du)/(dx)=d/dx[cos(1-2x)]=d/dx[cos(v)]$

$(du)/(dx)=(du)/(dv)*(dv)/(dx)$

$dy/dx=(dy)/(du)* (du)/(dv) *(dv)/(dx)$

$(du)/(dv)=-sin(v)$
$(dv)/(dx)=-2$

$dy/dx=2u*-sin(v)*-2$

$dy/dx=4usin(v)$

$dy/dx=4cos(1-2x)sin(1-2x)$

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