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

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Answer 1 · 2018-04-08T09:37:14

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

WE are given $y = arctan (cos x)$ $y=arctan(cosx)$

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

Then $\frac{d g (x)}{d x} = - sin x$ $(dg(x))/(dx)=-sinx$ and $y = arctan (g (x))$ $y=arctan(g(x))$

and according to chain rule $\frac{d y}{d x} = \frac{d y}{d g (x)} \cdot \frac{d g (x)}{d x}$ $(dy)/(dx)=(dy)/(dg(x))*(dg(x))/(dx)$

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

and as $g (x) = cos x$ $g(x)=cosx$ , $\frac{d g}{d x} = - sin x$ $(dg)/(dx)=-sinx$

Hence $\frac{d y}{d x} = \frac{1}{1 + {(g (x))}^{2}} \cdot (- sin x)$ $(dy)/(dx)=1/(1+(g(x))^2)*(-sinx)$

= $\frac{- sin x}{1 + {cos}^{2} x}$ $(-sinx)/(1+cos^2x)$

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