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Basic Inverse Trigonometric Functions

Solve y=tan...

Question

Solve $y = {tan}^{- 1} \frac{5 a x}{a^{2} - 6 x^{2}}$ , find $\frac{d y}{d x}$

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Solution

$y = {tan}^{- 1} (\frac{5 a x}{a^{2} - 6 x^{2}})$
$tan y = \frac{5 a x}{a^{2} - 6 x^{2}} * {sec}^{2} y - {tan}^{2} y = 1$
$({sec}^{2} y) \frac{d y}{d x} = \frac{5 a (a^{2} - 6 x^{2}) - (- 12 x) (5 a x)}{{(a^{2} - 6 x^{2})}^{2}}$
$[1 + {(\frac{5 a x}{a^{2} - 6 x^{2}})}^{2}] \frac{d y}{d x} = \frac{5 a^{3} - 30 a x^{2} + 60 a x^{2}}{{(a^{2} - 6 x^{2})}^{2}}$
$[\frac{a^{4} + 36 x^{4} + 13 a^{2} x^{2}}{{(a^{2} - 6 x^{2})}^{2}}] \frac{d y}{d x} = \frac{5 a^{3} + 30 a x^{2}}{{(a^{2} - 6 x^{2})}^{2}}$
$[(a^{4}) + 36 x^{4} + 13 a^{2} x^{2}] \frac{d y}{d x} = 5 a (a^{2} + 6 x^{2})$
$(a^{2} + 9 x^{2}) (a^{2} + 4 x^{2}) \frac{d y}{d x} = 5 a (a^{2} + 6 x^{2})$
$∴ \frac{d y}{d x} = \frac{5 a (a^{2} + 6 x^{2})}{(a^{2} + 9 x^{2}) (a^{2} + 4 x^{2})}$
$\frac{d y}{d x} = \frac{3 a (a^{2} + 4 x^{2}) + 2 a (a^{2} + 9 x^{2})}{(a^{2} + 9 x^{2}) (a^{2} + 4 x^{2})}$
$∴ \frac{d y}{d x} = \frac{3 a}{a^{2} + 9 x^{2}} + \frac{2 a}{a^{2} + 4 x^{2}}$

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y = {tan}^{- 1} (\frac{5 a x}{a^{2} - 6 x^{2}})

\frac{d y}{d x} = \frac{3 a}{a^{2} + 9 x^{2}} + \frac{2 a}{a^{2} + 4 x^{2}}

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