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Continuity & Differentiability

5.3. Derivatives of Inverse Trigonometric Functions.

Q.1. If y = tan –1[{√(1 + x²) – √(1 – x²)}/{√(1 + x²) + √(1 – x²)}] , find dy/dx.

Solution :

We have, y = tan –1[{√(1 + x²) – √(1 – x²)}/{√(1 + x²) + √(1 – x²)}]
Put x² = cos θ, then
y = tan –1[{√(1 + cos θ) – √(1 – cos θ)}/{√(1 + cos θ) + √(1 – cos θ)}]
= tan ^-1 [(√2 cos θ/2 – √2 sin θ/2)/(√2 cos θ/2 + √2 sin θ/2)]
= tan ^-1 [(1 – tan θ/2)/(1 + tan θ/2)]
= tan ^-1 [tan(π/4 – θ/2)]
= π/4 – θ/2 = π/4 – 1/2 cos –1 x².
Differentiating w.r.t. x we get
dy/dx = 0 – 1/2. d/dx(cos^-1 x²)
= – 1/2. – 1/√{1 – (x2)2}.d/dx(x²)
= 1/2 .1/√(1 – x⁴). 2x
= x/√(1 – x⁴). [Ans.]

Q.2. Differentiate the following with respect to x : tan –1[{√(1 + x) – √(1 – x)}/{√(1 + x) + √(1 – x)}]

Solution :

Let y = tan ^-1 [{√(1 + x) – √(1 – x)}/{√(1 + x) + √(1 – x)}]
= tan ^-1[{√(1 + cos θ) – √(1 – cos θ)}/{√(1 + cos θ) + √(1 – cos θ)}]
[Putting x = cos θ]
= tan^-1[{√(2cos2 θ/2) – √(2sin2 θ/2)}/{√(2cos2 θ/2) + √(2sin2 θ/2)}]
= tan ^-1[(cos θ/2 – sin θ/2)/(cos θ/2 + sin θ/2)]
= tan ^-1[(1 – tan θ/2)/(1 + tan θ/2)]
= tan ^-1[tan(π/4 – θ/2)]
= π/4 – θ/2
= π/4 – (1/2)cos ^-1x.
Therefore, dy/dx = 0 – (1/2)[–1/√(1 – x²)]
= 1/[2√(1 – x²)]. [Ans.]

Q.3. If y = cot ^-1[{√(1 + sin x) + √(1 – sin x)}/{√(1 + sin x) – √(1 – sin x)}], find dy/dx.

Solution :

We have, y = cot ^-1[{√(1 + sin x) + √(1 – sin x)}/{√(1 + sin x) – √(1 – sin x)}].
Let us consider, 1 + sin x = cos² x/2 + sin² x/2 + 2sin x/2.cos x/2
= (cos x/2 + sin x/2)²
Therefore, √(1 + sin x) = cos x/2 + sin x/2.
Similarly, √(1 – sin x) = cos x/2 – sin x/2.
Therefore, y = cot –1[{(cos x/2 + sin x/2) + (cos x/2 – sin x/2)}
/{(cos x/2 + sin x/2) – (cos x/2 – sin x/2)}]
= cot ^-1[(2 cos x/2)/(2 sin x/2)]
= cot ^-1[cot x/2]
= x/2.
Therefore, dy/dx = 1/2. [Ans.]

Paper By Mr. M.P.Keshari
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