How do you take the derivative of $y = sin x cos x - cos^2x$ ?

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Answer 1 · 2018-03-31T10:14:36

$(dy)/(dx)=cos2x+sin2x$

Here,
$y=sinxcosx-cos^2x$

$dy/dx=sinxd/(dx)(cosx)+cosxd/(dx)(sinx)-2cosxd/(dx)(cosx)$

$=sinx(-sinx)+cosxcosx-2cosx(-sinx)$

$=-sin^2x+cos^2x+2sinxcosx$

$=cos^2x-sin^2x+2sinxcosx$

$=cos2x+sin2x$

Note :

$(1) sin2x=2sinxcosx$

$(2)cos2x=cos^2x-sin^2x$

How do you take the derivative of y = sin x cos x - cos^2xy = sin x cos x - cos^2x?