If y-tan-13x-x31-3x2 then find dydx-

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Properties with Negative X

If y=tan-13...

Question

If $y = {tan}^{- 1} \frac{3 x - x^{3}}{1 - 3 x^{2}}$ then find $\frac{d y}{d x}$ .

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Solution

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y = {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}), - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}

\frac{d y}{d x}

\frac{d y}{d x},

y = {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}), \frac{- 1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}

y = {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}), - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} .

\frac{d y}{d x}

\frac{d y}{d x}

y = {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}), - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}

y = {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}})

\frac{d y}{d x}

x ϵ (- \infty, - \frac{1}{\sqrt{3}})

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