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Chain Rule of Differentiation

Differentiate...

Question

Differentiate $\sin^{- 1} (4 x \sqrt{1 - 4 x^{2}})$ with respect to $\sqrt{1 - 4 x^{2}}$ , if

(i) $x \in (- \frac{1}{2 \sqrt{2}}, \frac{1}{\sqrt{2 \sqrt{2}}})$

(ii) $x \in (\frac{1}{2 \sqrt{2}}, \frac{1}{2})$

(iii) $x \in (- \frac{1}{2}, - \frac{1}{2 \sqrt{2}})$

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Solution

$(i) Let, u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) put 2 x = \cos θ \Rightarrow u = \sin^{- 1} (2 \times \cos θ \sqrt{1 - \cos^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \cos θ \sin θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) Let, v = \sqrt{1 - 4 x^{2}} . . . (ii) Here, x \in (- \frac{1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}}) \Rightarrow 2 x \in (- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \Rightarrow θ \in (\frac{π}{4}, \frac{3 π}{4}) So, from equation (i), u = π - 2 θ [Since, \sin^{- 1} (sinθ) = π - θ, if θ \in (\frac{π}{2}, π)] \Rightarrow u = π - 2 \cos^{- 1} (2 x) [Since, 2 x = \cos θ]$

Differentiating it with respect to x,

$\frac{d u}{d x} = 0 - 2 (\frac{- 1}{\sqrt{1 - {(2 x)}^{2}}}) \frac{d}{d x} (2 x) \Rightarrow \frac{d u}{d x} = \frac{2}{\sqrt{1 - 4 x^{2}}} (2) \Rightarrow \frac{d u}{d x} = \frac{4}{\sqrt{1 - 4 x^{2}}} . . . (iii) from equation (ii) \frac{d v}{d x} = \frac{- 4 x}{\sqrt{1 - 4 x^{2}}} but, x \in (- \frac{1}{2}, - \frac{1}{2 \sqrt{2}}) \frac{d v}{d x} = \frac{- 4 (- x)}{\sqrt{1 - 4 {(- x)}^{2}}} \Rightarrow \frac{d v}{d x} = \frac{4 x}{\sqrt{1 - 4 x^{2}}} . . . (iv) Diferentiating equation (ii) with respect to x, \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} \frac{d}{d x} (1 - 4 x^{2}) \Rightarrow \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} (- 8 x) \Rightarrow \frac{d v}{d x} = \frac{- 4 x}{\sqrt{1 - 4 x^{2}}} . . . (v) Dividing equation (iii) by (v) \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{4}{\sqrt{1 - 4 x^{2}}} \times \frac{\sqrt{1 - 4 x^{2}}}{- 4 x} ∴ \frac{d u}{d v} = - \frac{1}{x}$

$(ii) Let, u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) put 2 x = \cos θ u = \sin^{- 1} (2 \times \cos θ \sqrt{1 - \cos^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \cos θ \sin θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) Let, v = \sqrt{1 - 4 x^{2}} . . . (ii) Here, x \in (\frac{1}{2 \sqrt{2}}, \frac{1}{2}) \Rightarrow 2 x \in (\frac{1}{\sqrt{2}}, 1) \Rightarrow \cos θ \in (\frac{1}{\sqrt{2}}, 1) \Rightarrow θ \in (0, \frac{π}{4}) So, from equation (i), u = 2 θ [Since, \sin^{- 1} (sinθ) = θ, if θ \in (- \frac{π}{2}, \frac{π}{2})] \Rightarrow u = 2 \cos^{- 1} (2 x) [Since, 2 x = \cos θ]$

Differentiate it with respect to x,

$\frac{d u}{d x} = 2 (\frac{- 1}{\sqrt{1 - {(2 x)}^{2}}}) \frac{d}{d x} (2 x) \frac{d u}{d x} = \frac{- 2}{\sqrt{1 - 4 x^{2}}} (2) \frac{d u}{d x} = \frac{- 4}{\sqrt{1 - 4 x^{2}}} . . . (iii) Diferentiating equation (ii) with respect to x, \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} \frac{d}{d x} (1 - 4 x^{2}) \Rightarrow \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} (- 8 x) \Rightarrow \frac{d v}{d x} = \frac{- 4 x}{\sqrt{1 - 4 x^{2}}} . . . (iv) Dividing equation (iii) by (iv) \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{- 4}{\sqrt{1 - 4 x^{2}}} \times \frac{\sqrt{1 - 4 x^{2}}}{- 4 x} ∴ \frac{d u}{d v} = \frac{1}{x}$

$(iii) Let, u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) put, 2 x = \cos θ \Rightarrow u = \sin^{- 1} (2 \times \cos θ \sqrt{1 - \cos^{2} θ}) \Rightarrow u = \sin^{- 1} (2 \cos θ \sin θ) \Rightarrow u = \sin^{- 1} (\sin 2 θ) . . . (i) Let, v = \sqrt{1 - 4 x^{2}} . . . (ii) Here, x \in (- \frac{1}{2}, - \frac{1}{2 \sqrt{2}}) \Rightarrow 2 x \in (- 1, - \frac{1}{\sqrt{2}}) \Rightarrow θ \in (\frac{3 π}{4}, π) So, from equation (i), u = π - 2 θ [Since, \sin^{- 1} (sinθ) = π - θ, if θ \in (\frac{π}{2}, \frac{3 π}{2})] \Rightarrow u = π - 2 \cos^{- 1} (2 x) [Since, 2 x = \cos θ]$

Differentiate it with respect to x,

$\frac{d u}{d x} = 0 - 2 (\frac{- 1}{\sqrt{1 - {(2 x)}^{2}}}) \frac{d}{d x} (2 x) \Rightarrow \frac{d u}{d x} = \frac{2}{\sqrt{1 - 4 x^{2}}} (2) \Rightarrow \frac{d u}{d x} = \frac{4}{\sqrt{1 - 4 x^{2}}} . . . (iii) from equation (ii), \frac{d v}{d x} = \frac{- 4 x}{\sqrt{1 - 4 x^{2}}} but, x \in (- \frac{1}{2}, - \frac{1}{2 \sqrt{2}}) ∴ \frac{d v}{d x} = \frac{- 4 (- x)}{\sqrt{1 - 4 {(- x)}^{2}}} \Rightarrow \frac{d v}{d x} = \frac{4 x}{\sqrt{1 - 4 x^{2}}} . . . (iv) Dividing equation (iii) by (iv) \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{4}{\sqrt{1 - 4 x^{2}}} \times \frac{\sqrt{1 - 4 x^{2}}}{4 x} ∴ \frac{d u}{d v} = \frac{1}{x}$

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Similar questions

Q. Differentiate

\sin^{- 1} (4 x \sqrt{1 - 4 x^{2}})

with respect to

\sqrt{1 - 4 x^{2}}

x \in (- \frac{1}{2 \sqrt{2}}, \frac{1}{\sqrt{2 \sqrt{2}}})

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