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Nature of Roots

Find dy/dx ...

Question

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

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Solution

$\frac{d}{d x} ({tan}^{- 1} \frac{5 x + 1}{3 - x - 6 x^{2}}) A p p l y i n g c h a i n r u l e \frac{d}{d x} f (u) = \frac{d f}{d u} \cdot \frac{d u}{d x} L e t u = \frac{5 x + 1}{3 - x - 6 x^{2}} = \frac{d}{d u} {tan}^{- 1} u \cdot \frac{d}{d x} (\frac{5 x + 1}{3 - x - 6 x^{2}}) W e k n o w t h a t \frac{d}{d u} {tan}^{- 1} u = \frac{1}{u^{2} + 1} S t e p 1 : \frac{d}{d u} {tan}^{- 1} u = \frac{1}{u^{2} + 1} S t e p 2 : \frac{d}{d x} (\frac{5 x + 1}{3 - x - 6 x^{2}}) = \frac{(5 x + 1) (3 - x - 6 x^{2}) - \frac{d}{d x} (3 - x - 6 x^{2}) (5 x + 1)}{{(3 - x - 6 x^{2})}^{2}} = \frac{5 (3 - x - 6 x^{2}) - (- 12 x - 1) (5 x + 1)}{{(3 - x - 6 x^{2})}^{2}} = \frac{30 x^{2} + 12 x + 16}{{(3 - x - 6 x^{2})}^{2}} T h e r e f o r e \frac{d}{d x} (\frac{5 x + 1}{3 - x - 6 x^{2}}) = \frac{30 x^{2} + 12 x + 16}{{(3 - x - 6 x^{2})}^{2}} (U s i n g q u o t i e n t r u l e o f d i f f e r e n t i a t i o n) C o m b i n i n g s t e p 1 a n d s t e p 2 : \frac{d}{d u} {tan}^{- 1} u = \frac{1}{u^{2} + 1} a n d \frac{d}{d x} (\frac{5 x + 1}{3 - x - 6 x^{2}}) = \frac{30 x^{2} + 12 x + 16}{{(3 - x - 6 x^{2})}^{2}} = \frac{1}{{[\frac{5 x + 1}{3 - x - 6 x^{2}}]}^{2} + 1} \cdot \frac{30 x^{2} + 12 x + 16}{{(3 - x - 6 x^{2})}^{2}} = \frac{15 x^{2} + 6 x + 8}{18 x^{4} + 6 x^{3} + 5 x^{2} + 2 x + 5} H e n c e \frac{d}{d x} ({tan}^{- 1} \frac{5 x + 1}{3 - x - 6 x^{2}}) = \frac{15 x^{2} + 6 x + 8}{18 x^{4} + 6 x^{3} + 5 x^{2} + 2 x + 5}$

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Similar questions

\frac{d y}{d x}

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

\frac{d y}{d x} i f y = {tan}^{- 1} [\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}]

y = {tan}^{- 1} \frac{5 x}{1 - 6 x^{2}}, - \frac{1}{\sqrt{6}} < x < \frac{1}{\sqrt{6}}

\frac{d y}{d x}

Q. Differentiate w.r.t.

x

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

$\frac{d}{d x} [t a n^{- 1} (\frac{5 x}{1 - 6 x^{2}})] =$

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