How do you find the derivative of $y= sin{cos^2(tanx)}$ ?

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Answer 1 · 2017-11-08T12:47:39

$(dy)/(dx)=-2sec^2(x)sin(tanx)cos(tanx)cos(cos^2(tan(x)))$

$y= sin(cos^2(tanx))=sin(f(x))$

$y=sin(f(x))=>(dy)/(dx)=f'(x)cos(f(x))$

$f(x)=cos^2(g(x))=(cos(g(x)))^2$
$f'(x)=2*d/(dx)[cos(g(x))] * cos(g(x))$
$f'(x)=2*-g'(x)sin(g(x))*cos(g(x))$
$f'(x)=-2g'(x)sin(g(x))cos(g(x))$

$g(x)=tanx$
$g'(x)=sec^2x$

Therefore, $f'(x)=-2sec^2(x)sin(tanx)cos(tanx)$

$(dy)/(dx)=-2sec^2(x)sin(tanx)cos(tanx)cos(cos^2(tan(x)))$

How do you find the derivative of y= sin{cos^2(tanx)}y= sin{cos^2(tanx)}?