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Differentiation of Inverse Trigonometric Functions

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Question

Find $\frac{d y}{d x}$ if $y = {tan}^{- 1} (\frac{5 x + 1}{3 - x - 6 x^{2}})$ .

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Solution

$y = {tan}^{- 1} \frac{5 x + 1}{3 - x - 6 x^{2}}$

$∴ tan y = \frac{5 x + 1}{3 - x - 6 x^{2}}$
differentiating wrt $x$ ,
${sec}^{2} y \frac{d y}{d x} = \frac{(3 - x - 6 x^{2}) \frac{d (5 x + 1)}{d x} - (5 x + 1) \frac{d (3 - x - 6 x^{2})}{d x}}{(3 - x - 6 x^{2})^{2}}$
$∴ (1 + {tan}^{2} y) \frac{d y}{d x} = \frac{5 (3 - x - 6 x^{2}) - (5 x + 1) (- 1 - 12 x)}{(3 - x - 6 x^{2})^{2}}$
$∴ (1 + {[\frac{5 x + 1}{3 - x - 6 x^{2}}]}^{2}) \frac{d y}{d x} = \frac{15 - 5 x - 30 x^{2} + 5 x + 60 x^{2} + 1 + 12 x}{(3 - x - 6 x^{2})^{2}}$
$∴ (\frac{(3 - x - 6 x^{2})^{2} + (5 x + 1)^{2}}{(3 - x - 6 x^{2})^{2}}) \frac{d y}{d x} = \frac{30 x^{2} + 12 x + 16}{(3 - x - 6 x^{2})^{2}}$
$∴ \frac{d y}{d x} = \frac{30 x^{2} + 12 x + 16}{(3 - x - 6 x^{2})^{2} + (5 x + 1)^{2}}$
$∴ \frac{d y}{d x} = \frac{30 x^{2} + 12 x + 16}{10 + 4 x - 10 x^{2} + 12 x^{3} + 36 x^{4}}$
$∴ \frac{d y}{d x} = \frac{15 x^{2} + 6 x + 8}{5 + 2 x - 5 x^{2} + 6 x^{3} + 18 x^{4}}$
This is the required answer as the denominator has no roots.

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