第12类RD Sharma解决方案-第11章区分–练习11.8 |套装2

问题11：如果-1 /√2-1 (x /√(1-x ^{2 ))区分sin -1} (2x√(1-x ^2))。

解决方案：

Let u = sin^-1(2x√(1-x²))

Substitute x = sin θ ⇒ θ = sin^-1x

u = sin^-1(2 sin θ √(1 – sin²θ))

u = sin^-1(2 sin θ cos θ) [sin²θ + cos²θ = 1]

u = sin^-1(sin 2θ) —–(i) [sin 2θ = 2 sin θ cos θ]

Let, v = tan^-1(x/√(1-x²))

v = tan^-1(sin θ/ √(1 – sin²θ))

v = tan^-1(sin θ/cosθ) [sin²θ + cos²θ = 1]

v = tan^-1(tan θ) —–(ii) [tan θ = sin θ/cos θ]

Here, -1/√2 < x <1/√2

⇒ -1/√2 < sin θ <1/√2

⇒ – π/4 < θ < π/4

So, from equation (i), u = 2 θ [sin^-1(sin θ) = θ, θ ∈ (0,π/2)]

u = 2 sin^-1x

du/dx = 2/ √(1 – x²) —–(iii) [d(sin^-1x)/dx = 1/√(1 – x²)]

From equation (ii), v = θ [tan^-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

v = sin^-1x

dv/dx = 1/ √(1 – x²) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 2 (Ans)

问题12：如果0 -1 ((1-x ² )/(1 + x ² ))区分tan ^-1 (2x /(1-x ²⁾⁾

解决方案：

Let u = tan^-1(2x/(1 – x²))

Substitute x = tan θ, ⇒ θ = tan^-1x

u = tan^-1(2 tan θ/(1 – tan²θ))

u = tan^-1(tan 2θ) —–(i) [tan 2θ = 2 tan θ/(1 – tan²θ)]

Now, let v = cos^-1((1 – tan²θ)/(1 + tan²θ))

v = cos^-1(cos 2θ) —–(ii) [cos 2θ = (1 – tan²θ)/(1 + tan²θ)]

Here, 0 < x <1

⇒ 0 < tan θ < 1

⇒ 0 < θ < π/4

So, from equation (i), u = 2θ [tan^-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

u = 2 tan^-1x

du/dx = 2/(1 + x²) —–(iii) [d(tan^-1x)/dx = 1/(1 + x²)]

From equation (ii), v = 2θ [cos^-1(cos θ) = θ, θ ∈ [0 , π]

v = 2 tan^-1x

dv/dx = 2/(1 + x²) —–(iv) [d(tan^-1x)/dx = 1/(1 + x²)]

Dividing equation (iii) by (iv)

du/dv = 1 (Ans)

问题13：如果-1/2 ，则相对于sin -1 (3x-4x ³ )区分tan ^-1 ((x-1)/(x + 1))。

解决方案：

Let u = tan^-1((x – 1)/(x + 1))

Substitute x = tan θ ⇒ θ = tan^-1x

u = tan^-1((tan θ – 1)/(tan θ + 1))

u = tan^-1((tan θ – tan π/4)/(1 + tan θ tan π/4))

u = tan^-1(tan(θ – π/4) —–(i) [tan(A – B) = (tan A – tan B)/(1 + tan A * tan B]

Here, -1/2 < x < 1/2

⇒ -1/2 < tan θ < 1/2

⇒ -tan^-1(1/2) < θ < tan^-1(1/2)

So, from equation (i), u = θ – π/4 [tan^-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

u = tan^-1x – π/4

du/dx = 1/(1 + x²) —–(ii) [d(tan^-1x)/dx = 1/(1 + x²)]

Now, let v = sin^-1(3x – 4x³)

Substitute x = sin θ

v = sin^-1(3 sin θ – 4 sin³θ)

v = sin^-1(sin 3θ) —–(iii) [sin 3θ = 3 sin θ – 4 sin³θ]

Here -1/2

⇒ -1/2 < sin θ < 1/2

⇒ – π/6 < θ < π/6

From equation (ii), v = 3θ [sin^-1(sin θ) = θ, θ ∈ (-π/2 , π/2)]

v = 3 sin^-1x

dv/dx = 3/ √(1 – x²) —–(iv) [d(sin^-1x)/dx = 1/√(1 – x²)]

Dividing equation (ii) by (iv)

du/dv = (1/3) * √(1 – x²)/(1 + x²) (Ans)

问题14：^{关于sec -1} x区分出tan ^-1 (cosx /(1 + sinx))。

解决方案：

Let u = tan^-1(cos x/(1 + sin x))

u = tan^-1((cos²(x/2) – sin²(x/2))/(cos²(x/2) + sin²(x/2) + 2sin(x/2)cos(x/2)) [cos 2x = cos²x – sin²x , cos²x + sin²x = 1, sin 2x = 2sin x cos x]

Using [a² – b² =(a-b)(a+b) ,and (a+b)² = a² + b² + 2ab],

u = tan^-1(((cos(x/2) – sin(x/2)) * (cos(x/2) + sin(x/2)))/(cos(x/2) + sin(x/2))²) 

u = tan^-1((cos(x/2) – sin(x/2))/(cos(x/2 + sin(x/2)))

Dividing numerator and denominator by cos(x/2)

u = tan^-1((1 – tan(x/2))/(1 + tan(x/2)))

u = tan^-1((tan π/4 – tan(x/2))/(1 + tan(x/2)*tan π/4))

u = tan^-1(tan(π/4 – x/2) [tan(A – B) = (tan A – tan B)/(1 + tan A * tan B]

u = π/4 – x/2

du/dx = -1/2 —–(i)

Now, let v = sec^-1x

dv/dx = 1/(x√(x²-1)) —–(ii)

Dividing equation (i) by (ii)

du/dv = ½ * (- x√(x²-1)) (Ans.)

问题15：如果-1 -1 (2x /(1-x ² ))区分sin ^-1 (2x /(1 + x ^2))。

解决方案：

Let u = sin^-1(2x/(1+x²))

Substitute x = tan θ ⇒ θ = tan^-1x

u = sin^-1((2 tanθ/ (1 + tan²θ))

u = sin^-1(sin 2θ) —–(i) [sin2θ = 2 tanθ/(1 + tan²θ)]

Now, let v = tan^-1(2x/(1 – x²))

v = tan^-1(2 tan θ/(1 – tan²θ))

v = tan^-1(tan 2θ) —–(ii) [tan 2θ = 2 tan θ/(1 – tan²θ)]

Here, -1

⇒ -1

⇒ – π/4 < θ < π/4

So, from equation (i), u = 2θ [sin^-1(sin θ) = θ, θ ∈ (-π/2 , π/2)]

u = 2tan^-1x

du/dx = 2/(1 + x²) —–(iii)

From equation (ii), v = 2θ [tan^-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

v = 2tan^-1x

dv/dx = 2/(1+x²) ——(iv)

Dividing equation (iii) by (iv)

du/dv = 1 (Ans)

问题16：如果1/2 -1 (√(1-x ² )/ x)区分cos ^-1 (4x ^{3 -3x)。}

解决方案：

Let u = cos^-1(4x³-3x)

Substitute x = cos θ ⇒ θ = cos^-1x

u = cos^-1(4 cos³θ – 3cos θ)

u = cos^-1(cos3θ) —–(i) [cos 3θ = 4 cos³θ – 3cos θ]

Now, let v = tan^-1(√(1 – x²)/x)

v = tan^-1(√(1 – cos²θ)/cos θ)

v = tan^-1(sin θ/cos θ)

v = tan^-1(tan θ) —–(ii)

Here, 1/2 < x < 1

⇒ 1/2 < cos θ < 1

⇒ 0 < θ < π/3

From equation (i), u = 3θ [cos^-1(cos θ) = θ, θ ∈ [0 , π]

u = 3cos^-1x

du/dx = -3/√(1-x²) —–(iii)

From equation (ii), v = θ [tan^-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

v = cos^-1x

dv/dx = -1/√(1-x2) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 3 (Ans)

问题17：如果-1 /√2-1 (2x√(1-x ² ))区分tan ^-1 (x /√(1-x ^2))。

解决方案：

Let u = tan^-1(x/√(1-x²))

Substitute x = sin θ ⇒ θ = sin^-1x

u = tan^-1(sinθ/√(1 – sin²θ))

u = tan^-1(sin θ/cos θ)

u = tan^-1(tan θ) —–(i)

And, v = sin^-1(2x√(1-x²))

v = sin^-1(2 sin θ √(1-sin²θ))

v = sin^-1(2 sin θ cos θ)

v = sin^-1(sin 2θ) —–(ii)

Here, -1/√2

⇒ -1/√2 < sin θ < 1/√2

⇒ – π/4 < θ < π/4

From equation (i), u = θ [tan^-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

u = sin^-1x

du/dx = 1/ √(1-x²) —–(iii)

From equation (ii), v = 2θ [sin^-1(sin θ) = θ, θ ∈ (-π/2, π/2)]

v = 2sin^-1x

dv/dx = 2/√(1-x²) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 1/2 (Ans)

问题18：微分罪^{-1√(1-x 2)}相对于床^-1(X /√(1-x ^2))，如果0

解决方案：

Let u = sin^-1√(1 – x² )

Substitute x = cos θ ⇒ θ = cos^-1x

u = sin^-1√(1 – cos²θ)

u = sin^-1(sin θ) —–(i)

And, v = cot^-1(x/√(1-x²))

v = cot^-1(cos θ/√(1 – cos²θ))

v = cot^-1(cos θ/sin θ)

v = cot ^-1(cot θ)

v = tan^-1(tan θ) —–(ii) [tan^-1(θ) = cot^-1(1/θ)]

Here, 0 < x <1

0 < cos θ <1

0 < θ < π/2

So, from equation (i), u = θ [sin^-1(sin θ) = θ, θ ∈ (-π/2 , π/2)]

u = cos^-1x

du/dx = -1/ √(1-x²) —–(iii)

And, from equation (ii), v = θ [tan^-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

v = cos^-1x

dv/dx = -1/ √(1-x²) —–(iv)

Dividing equation (iii) by (iv)

du/dv = 1 (Ans)

问题19：如果-1 /√22 x ² )区分sin ^-1 (2ax√(1-a ² x ²⁾⁾

解决方案：

Let u = sin^-1(2ax√(1-a²x²))

Substitute ax = sin θ ⇒ θ = sin^-1(ax)

u = sin^-1(2sin θ √(1 – sin²θ))

u = sin^-1(2 sin θ cos θ)

u = sin^-1(sin 2θ) —–(i)

And, let v = √(1-a²x²)

dv/dx = -a²x/ √(1 – a²x²) —–(ii)

Here, -1/√2

⇒ -1/√2 < sin θ <1/√2

⇒ – π/4 < θ < π/4

So, from equation (i), u = 2θ [sin^-1(sin θ) = θ, θ ∈ (-π/2 , π/2)]

u = 2sin^-1(ax)

du/dx = 2a/ √(1 – a²x²) —–(iii)

Dividing equation (iii) by (ii)

du/dv = -2/ax (Ans)

问题20：如果-1 ，则相对于√(1-x 2 ^{)区分tan -1} ((1-x)/(1 + x))。

解决方案：

Let u = tan^-1((1-x)/(1+x))

Substitute x = tan θ ⇒ θ = tan^-1x

u = tan^-1((1 – tan θ)/(1 + tan θ))

u = tan^-1((tan π/4 – tan θ)/(1 + tan θ tan π/4))

u = tan^-1(tan(π/4 – θ) —–(i) [tan(A – B) = (tan A – tan B)/(1 + tan A * tan B]

Here, -1 < x < 1

⇒ -1 < tan θ < 1

⇒ – π/4 < θ < π/4

So, from equation (i), u = π/4 – θ [tan^-1(tan θ) = θ, θ ∈ [- π/2, π/2]]

u = π/4 – tan^-1x

du/dx = -1/(1 + x²) —–(ii)

And, v = √(1 – x²)

dv/dx = -2x/(2 * √(1 – x²)) = – x/√(1 – x²) —–(iii)

Dividing equation (ii) by (iii)

du/dv = √(1 – x²)/(x * (1 + x²)) (Ans)