Math, asked by BrainlyHelper, 1 year ago

Find the second order derivatives of the function.tan^-1 x

Answers

Answered by abhi178
9
Let \bf{y=tan^{-1}x}
now differentiate y with respect to x ,
\bf{\frac{dy}{dx}=\frac{d(tan^{-1}x)}{dx}}\\\\=\bf{\frac{1}{1+x^2}}

so, \bf{\frac{dy}{dx}=\frac{1}{1+x^2}}

now differentiate \frac{dy}{dx} once again.

\bf{\frac{d^2y}{dx^2}=\frac{d(1+x^2)^{-1}}{dx}}\\\\=\bf{-(1+x^2)^{-1-1}(2x)}\\\\=\bf{\frac{-2x}{(1+x^2)^2}}

hence, d²y/dx² = -2x/(1 + x²)²
Answered by rohitkumargupta
9
let y = tan^-1 x

now differentiate with respect to x.

dy/dx = \bold{1/(x^2 + 1)}

differentiating once again,

d²y/dx² = \bold{\frac{(x^2 + 1) * 0 - 1(2x + 0)}{(x^2 + 1)^2}}

d²y/dx² = \bold{\frac{-2x}{(x^2 + 1)^2}}
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