Differentiate sin-1(2x1+x2)w.r.t.cos-1(1-x21+x2) - Mathematics and Statistics

प्रश्न

Differentiate `sin^-1((2x)/(1 + x^2))w.r.t. cos^-1((1 - x^2)/(1 + x^2))`

बेरीज

उत्तर

Let u = `sin^-1((2x)/(1 + x^2))` and

v = `cos^-1((1 - x^2)/(1 + x^2))`

Then we want to find `"du"/"dv"`.
Put x = tanθ.
Then θ = tan^–1x.
u = `sin^-1((2tanθ)/(1 + tanθ))`
= sin^–1(sin2θ)
= 2θ
= 2tan^–1x
∴ `"du"/"dx" = 2"d"/"dx"(tan^-1x)`

= `2 xx (1)/(1 + x^2)`

= `(2)/(1 + x^2)`
Also, v = `cos^-1((1 - tan^2θ)/(1 + tan^2θ))`
= cos^–1(cos 2θ)
= 2θ
= 2 tan^–1x
∴ `"dv"/"dx" = 2"d"/"dx"(tan^-1x)`

= `2 xx (1)/(1 + x^2)`

= `(2)/(1 + x^2)`

∴ `"du"/"dv" = (("du"/"dx"))/(("dv"/"dx")`

= `(((2)/(1 + x^2)))/(((2)/(1 + x^2))`
= 1.
Alternative Method :
Let u = `sin^-1((2x)/(1 + x^2)) and v = cos^-1((1 - x^2)/(1 + x^2))`

Then we want to find `"du"/"dv"`
Put x = tanθ.
Then u = `sin^-1((2tanθ)/(1 + tanθ))`
= sin^–1 (sin2θ)
= 2θ
and
v = `cos^-1((1 - tan^2θ)/(1 + tan^2θ))`
= cos^–1 (cos2θ)
= 2θ
∴ u = v
Differentiating both sides w.r.t. v, we get
`"du"/"dv"` = 1.

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Derivatives of Implicit Functions

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 4.2 Q 4.1 Q 4.3

पाठ 1: Differentiation - Exercise 1.4 [पृष्ठ ४९]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board

पाठ 1 Differentiation
Exercise 1.4 | Q 4.2 | पृष्ठ ४९

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