Differentiate the following w.r.t. x : `tan^-1((sqrt(x)(3 - x))/(1 - 3x))`

Let y = `tan^-1((sqrt(x)(3 - x))/(1 - 3x))` = `tan^-1[(3sqrt(x) - xsqrt(x))/(1 - 3x)]` Put `sqrt(x) = tanθ`. Thenθ = `tan^-1(sqrt(x))` &there4; y = `tan^-1((3tanθ - tan^3θ)/(1 - 3tan^2θ))`= tan–1 (tan3θ)= 3θ= `3tan^-1(sqrt(x))` &there4; `"dy"/"dx" = 3"d"/"dx"[tan^-1(sqrt(x))]` = `3 xx (1)/(1 + (sqrt(x))^2)."d"/"dx"(sqrt(x))` = `(3)/(1 + x) xx (1)/(2sqrt(x))` = `(3)/(2sqrt(x)(1 + x)`.

Differentiate the following w.r.t. x : tan-1(x(3-x)1-3x) - Mathematics and Statistics

प्रश्न

Differentiate the following w.r.t. x : $\tan^{- 1} (\frac{\sqrt{x} (3 - x)}{1 - 3 x})$

बेरीज

उत्तर

Let y = $\tan^{- 1} (\frac{\sqrt{x} (3 - x)}{1 - 3 x})$

= $\tan^{- 1} [\frac{3 \sqrt{x} - x \sqrt{x}}{1 - 3 x}]$

Put $\sqrt{x} = \tan θ$ . Thenθ = $\tan^{- 1} (\sqrt{x})$

∴ y = $\tan^{- 1} (\frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ})$
= tan^–1 (tan3θ)
= 3θ
= $3 \tan^{- 1} (\sqrt{x})$

∴ $\frac{dy}{dx} = 3 \frac{d}{dx} [\tan^{- 1} (\sqrt{x})]$

= $3 \times \frac{1}{1 + {(\sqrt{x})}^{2}} . \frac{d}{dx} (\sqrt{x})$

= $\frac{3}{1 + x} \times \frac{1}{2 \sqrt{x}}$

= $\frac{3}{2 \sqrt{x} (1 + x)}$ .

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Derivatives of Implicit Functions

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Q 4.3 Q 4.2 Q 4.4

पाठ 1: Differentiation - Miscellaneous Exercise 1 (II) [पृष्ठ ६४]

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पाठ 1 Differentiation
Miscellaneous Exercise 1 (II) | Q 4.3 | पृष्ठ ६४

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Differentiate the following w.r.t. x : tan-1(x(3-x)1-3x) - Mathematics and Statistics

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Differentiate the following w.r.t. x : tan-1(x(3-x)1-3x) - Mathematics and Statistics

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