Question

find derivative.

​

Answer 1

$\boxed{ \red{\bold{- \frac{ 2}{1 + {x}^{2} } }}}$

◆In this question , first solve the Inverse Trigonometry,

by Let

$\begin{gathered}x = tan \theta \\ or \: \\ \theta = {tan}^{ - 1} x\end{gathered}$

●After solving , we get the value:-

$\boxed{- 2 {tan}^{ - 1} x}$

◆ And at last differentiate w.r.t X

Formula used:-

$\star\boxed{ \bold{ \pink{cos \: 2 \theta = \frac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta } }}}$

$\star \boxed{ \purple{\bold{ {cos}^{ - 1} (cosx) = x}}}$

$\star \: \boxed{ \pink{\bold{ \frac{d}{dx} ( {tan}^{ - 1} x) = \frac{1}{1 + {x}^{2} } }}}$

Solution refer to the attachment

Answer 2

Answer:

\boxed{ \red{\bold{- \frac{ 2}{1 + {x}^{2} } }}}

−

1+x

2

2

◆In this question , first solve the Inverse Trigonometry,

by Let

\begin{gathered}\begin{gathered}x = tan \theta \\ or \: \\ \theta = {tan}^{ - 1} x\end{gathered}\end{gathered}

x=tanθ

or

θ=tan

−1

x

●After solving , we get the value:-

\boxed{- 2 {tan}^{ - 1} x}

−2tan

−1

x

◆ And at last differentiate w.r.t X

Formula used:-

\star\boxed{ \bold{ \pink{cos \: 2 \theta = \frac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta } }}}⋆

cos2θ=

1+tan

2

θ

1−tan

2

θ

\star \boxed{ \purple{\bold{ {cos}^{ - 1} (cosx) = x}}}⋆

cos

−1

(cosx)=x

\star \: \boxed{ \pink{\bold{ \frac{d}{dx} ( {tan}^{ - 1} x) = \frac{1}{1 + {x}^{2} } }}}⋆

dx

d

(tan

−1

x)=

1+x

2

1

Solution refer to the attachment