"Derivative of sec-111-2x2 with respect to sin-13x-4x3 is." was answered by our community

leshumict6

Answered question

2023-02-10

Derivative of

\sec^{- 1} (\frac{1}{1 - 2 x^{2}})

with respect to

\sin^{- 1} (3 x - 4 x^{3})

is.

Izabella James

Beginner2023-02-11Added 3 answers

\frac{2}{3}

Explanation for correct option:
Differentiate

\sec^{- 1} (\frac{1}{1 - 2 x^{2}})

with respect to

x

.
Let,

u = \sec^{- 1} (\frac{1}{1 - 2 x^{2}})

Put,

x = \sin (θ) θ = \sin^{- 1} (x)

\Rightarrow u = \sec^{- 1} (\frac{1}{1 - 2 {(\sin (θ))}^{2}}) \Rightarrow u = \sec^{- 1} (\frac{1}{\cos (2 θ)}) (∵ 1 - 2 {(\sin (θ))}^{2} = \cos (2 θ)) \Rightarrow u = \sec^{- 1} (\sec (2 θ)) \Rightarrow u = 2 θ \Rightarrow u = 2 \sin^{- 1} (x)

Differentiate with respect to

x

\Rightarrow \frac{d u}{d x} = \frac{2}{1} . . . . . . . . . . . . . . . (i)

Differentiate

\sin^{- 1} (3 x - 4 x^{3})

with respect to

x

.
Let,

v = \sin^{- 1} (3 x - 4 x^{3})

Put,

x = \sin (θ) θ = \sin^{- 1} (x)

\Rightarrow v = \sin^{- 1} (3 \sin (θ) - 4 {(\sin (θ))}^{3}) \Rightarrow v = \sin^{- 1} (\sin (3 θ)) (∵ 3 \sin (θ) - 4 {(\sin (θ))}^{3} = \sin (3 θ)) \Rightarrow v = 3 θ \Rightarrow v = 3 \sin^{- 1} (x)

Differentiate with respect to

x

\Rightarrow \frac{d v}{d x} = \frac{3}{1} . . . . . . . . . . . . . . . (i i)

Divide the equation

(i)

(i i)

\Rightarrow \frac{(\frac{d u}{d x})}{(\frac{d v}{d x})} = \frac{\frac{2}{1}}{\frac{3}{1}} \Rightarrow \frac{d u}{d v} = \frac{2}{3}

Derivative of

\sec^{- 1} (\frac{1}{1 - 2 x^{2}})

with respect to

\sin^{- 1} (3 x - 4 x^{3})

\frac{2}{3}

.
Hence, correct answer is option

(C)

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