How do you find the derivative of the function: $y = arctan(cosx)$ ?

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Answer 1 · 2018-04-08T09:37:14

$(dy)/(dx)=(-sinx)/(1+cos^2x)$

WE are given $y=arctan(cosx)$

Let $y=f(x)=arctanx$ and $g(x)=cosx$

Then $(dg(x))/(dx)=-sinx$ and $y=arctan(g(x))$

and according to chain rule $(dy)/(dx)=(dy)/(dg(x))*(dg(x))/(dx)$

Now as $y=arctan(g(x))$ , we have $(dy)/(dg(x))=1/(1+(g(x))^2)$

and as $g(x)=cosx$ , $(dg)/(dx)=-sinx$

Hence $(dy)/(dx)=1/(1+(g(x))^2)*(-sinx)$

= $(-sinx)/(1+cos^2x)$

How do you find the derivative of the function: y = arctan(cosx) y = arctan(cosx)?