The question
(a) If 1 − x 2 + 1 − y 2 = a ( x − y ) \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) 1 − x 2 + 1 − y 2 = a ( x − y ) d y d x = 1 − y 2 1 − x 2 \dfrac{dy}{dx} = \sqrt{\dfrac{1 - y^2}{1 - x^2}} d x d y = 1 − x 2 1 − y 2
OR
(b) If y = ( tan x ) x y = (\tan x)^x y = ( tan x ) x d y d x \dfrac{dy}{dx} d x d y
Exam answer Show all explanations
(a)
Put x = sin θ x = \sin\theta x = sin θ y = sin ϕ y = \sin\phi y = sin ϕ
cos θ + cos ϕ = a ( sin θ − sin ϕ ) \cos\theta + \cos\phi = a(\sin\theta - \sin\phi) cos θ + cos ϕ = a ( sin θ − sin ϕ ) Why?
The roots 1 − x 2 \sqrt{1 - x^2} 1 − x 2 1 − y 2 \sqrt{1 - y^2} 1 − y 2 1 − sin 2 θ = cos 2 θ 1 - \sin^2\theta = \cos^2\theta 1 − sin 2 θ = cos 2 θ
With θ , ϕ \theta, \phi θ , ϕ [ − π 2 , π 2 ] \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] [ − 2 π , 2 π ] 1 − x 2 = cos θ \sqrt{1 - x^2} = \cos\theta 1 − x 2 = cos θ 1 − y 2 = cos ϕ \sqrt{1 - y^2} = \cos\phi 1 − y 2 = cos ϕ
1 − x 2 = 1 − sin 2 θ = cos 2 θ = cos θ 1 − y 2 = 1 − sin 2 ϕ = cos 2 ϕ = cos ϕ a ( x − y ) = a ( sin θ − sin ϕ ) \begin{aligned} \sqrt{1 - x^2} &= \sqrt{1 - \sin^2\theta} \\ &= \sqrt{\cos^2\theta} = \cos\theta \\[4pt] \sqrt{1 - y^2} &= \sqrt{1 - \sin^2\phi} \\ &= \sqrt{\cos^2\phi} = \cos\phi \\[4pt] a(x - y) &= a(\sin\theta - \sin\phi) \end{aligned} 1 − x 2 1 − y 2 a ( x − y ) = 1 − sin 2 θ = cos 2 θ = cos θ = 1 − sin 2 ϕ = cos 2 ϕ = cos ϕ = a ( sin θ − sin ϕ ) Using the sum-to-product formulas,
2 cos θ + ϕ 2 cos θ − ϕ 2 = 2 a cos θ + ϕ 2 sin θ − ϕ 2 \begin{aligned} &2\cos\frac{\theta + \phi}{2}\cos\frac{\theta - \phi}{2} \\ &= 2a\cos\frac{\theta + \phi}{2}\sin\frac{\theta - \phi}{2} \end{aligned} 2 cos 2 θ + ϕ cos 2 θ − ϕ = 2 a cos 2 θ + ϕ sin 2 θ − ϕ Why?
The two formulas used are:
cos C + cos D = 2 cos C + D 2 cos C − D 2 \begin{aligned} &\cos C + \cos D \\ &= 2\cos\frac{C + D}{2}\cos\frac{C - D}{2} \end{aligned} cos C + cos D = 2 cos 2 C + D cos 2 C − D sin C − sin D = 2 cos C + D 2 sin C − D 2 \begin{aligned} &\sin C - \sin D \\ &= 2\cos\frac{C + D}{2}\sin\frac{C - D}{2} \end{aligned} sin C − sin D = 2 cos 2 C + D sin 2 C − D Use them with C = θ C = \theta C = θ D = ϕ D = \phi D = ϕ a a a
a ( sin θ − sin ϕ ) = a ⋅ 2 cos θ + ϕ 2 sin θ − ϕ 2 = 2 a cos θ + ϕ 2 sin θ − ϕ 2 \begin{aligned} &a(\sin\theta - \sin\phi) \\ &= a \cdot 2\cos\frac{\theta + \phi}{2}\sin\frac{\theta - \phi}{2} \\ &= 2a\cos\frac{\theta + \phi}{2}\sin\frac{\theta - \phi}{2} \end{aligned} a ( sin θ − sin ϕ ) = a ⋅ 2 cos 2 θ + ϕ sin 2 θ − ϕ = 2 a cos 2 θ + ϕ sin 2 θ − ϕ Both sides now share the factor 2 cos θ + ϕ 2 2\cos\frac{\theta + \phi}{2} 2 cos 2 θ + ϕ
Dividing both sides by 2 cos θ + ϕ 2 sin θ − ϕ 2 2\cos\frac{\theta + \phi}{2}\sin\frac{\theta - \phi}{2} 2 cos 2 θ + ϕ sin 2 θ − ϕ
cot θ − ϕ 2 = a ⇒ θ − ϕ = 2 cot − 1 a ⇒ sin − 1 x − sin − 1 y = 2 cot − 1 a \begin{aligned} &\cot\frac{\theta - \phi}{2} = a \\ \Rightarrow\ &\theta - \phi = 2\cot^{-1}a \\ \Rightarrow\ &\sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a \end{aligned} ⇒ ⇒ cot 2 θ − ϕ = a θ − ϕ = 2 cot − 1 a sin − 1 x − sin − 1 y = 2 cot − 1 a Why?
The common factor cancels, and cos α sin α = cot α \frac{\cos\alpha}{\sin\alpha} = \cot\alpha s i n α c o s α = cot α α = θ − ϕ 2 \alpha = \frac{\theta - \phi}{2} α = 2 θ − ϕ
Left side divided:
2 cos θ + ϕ 2 cos θ − ϕ 2 2 cos θ + ϕ 2 sin θ − ϕ 2 = cos θ − ϕ 2 sin θ − ϕ 2 = cot θ − ϕ 2 \begin{aligned} &\frac{2\cos\frac{\theta + \phi}{2}\cos\frac{\theta - \phi}{2}}{2\cos\frac{\theta + \phi}{2}\sin\frac{\theta - \phi}{2}} \\ &= \frac{\cos\frac{\theta - \phi}{2}}{\sin\frac{\theta - \phi}{2}} = \cot\frac{\theta - \phi}{2} \end{aligned} 2 cos 2 θ + ϕ sin 2 θ − ϕ 2 cos 2 θ + ϕ cos 2 θ − ϕ = sin 2 θ − ϕ cos 2 θ − ϕ = cot 2 θ − ϕ Right side divided:
2 a cos θ + ϕ 2 sin θ − ϕ 2 2 cos θ + ϕ 2 sin θ − ϕ 2 = a \frac{2a\cos\frac{\theta + \phi}{2}\sin\frac{\theta - \phi}{2}}{2\cos\frac{\theta + \phi}{2}\sin\frac{\theta - \phi}{2}} = a 2 cos 2 θ + ϕ sin 2 θ − ϕ 2 a cos 2 θ + ϕ sin 2 θ − ϕ = a Taking cot − 1 \cot^{-1} cot − 1 θ − ϕ \theta - \phi θ − ϕ
cot θ − ϕ 2 = a θ − ϕ 2 = cot − 1 a θ − ϕ = 2 cot − 1 a ( × 2 ) \begin{aligned} \cot\frac{\theta - \phi}{2} &= a \\ \frac{\theta - \phi}{2} &= \cot^{-1}a \\ \theta - \phi &= 2\cot^{-1}a \quad (\times 2) \end{aligned} cot 2 θ − ϕ 2 θ − ϕ θ − ϕ = a = cot − 1 a = 2 cot − 1 a ( × 2 ) Then go back to x x x y y y θ = sin − 1 x \theta = \sin^{-1}x θ = sin − 1 x ϕ = sin − 1 y \phi = \sin^{-1}y ϕ = sin − 1 y
The right side, 2 cot − 1 a 2\cot^{-1}a 2 cot − 1 a
Differentiating both sides w.r.t. x x x
1 1 − x 2 − 1 1 − y 2 d y d x = 0 \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}}\,\frac{dy}{dx} = 0 1 − x 2 1 − 1 − y 2 1 d x d y = 0 Why?
The standard derivative is
d d x sin − 1 x = 1 1 − x 2 \frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}} d x d sin − 1 x = 1 − x 2 1 y is a function of x, so sin − 1 y \sin^{-1}y sin − 1 y 1 1 − y 2 ⋅ d y d x \frac{1}{\sqrt{1 - y^2}}\cdot\frac{dy}{dx} 1 − y 2 1 ⋅ d x d y
The derivative of the constant 2 cot − 1 a 2\cot^{-1}a 2 cot − 1 a
⇒ d y d x = 1 − y 2 1 − x 2 \Rightarrow \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} ⇒ d x d y = 1 − x 2 1 − y 2 Why?
Move the d y d x \frac{dy}{dx} d x d y 1 − y 2 \sqrt{1 - y^2} 1 − y 2
1 1 − x 2 = 1 1 − y 2 d y d x 1 − y 2 × 1 1 − x 2 = d y d x d y d x = 1 − y 2 1 − x 2 \begin{aligned} \frac{1}{\sqrt{1 - x^2}} &= \frac{1}{\sqrt{1 - y^2}}\,\frac{dy}{dx} \\[4pt] \sqrt{1 - y^2} \times \frac{1}{\sqrt{1 - x^2}} &= \frac{dy}{dx} \\[4pt] \frac{dy}{dx} &= \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \end{aligned} 1 − x 2 1 1 − y 2 × 1 − x 2 1 d x d y = 1 − y 2 1 d x d y = d x d y = 1 − x 2 1 − y 2 Both roots together: p q = p q \frac{\sqrt p}{\sqrt q} = \sqrt{\frac{p}{q}} q p = q p
∴ d y d x = 1 − y 2 1 − x 2 \therefore\ \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} ∴ d x d y = 1 − x 2 1 − y 2
(b)
y = ( tan x ) x y = (\tan x)^x y = ( tan x ) x Taking log on both sides,
log y = x log ( tan x ) \log y = x\log(\tan x) log y = x log ( tan x ) Why?
Here both the base and the power contain x x x
Taking log brings the power down in front, using
log m n = n log m \log m^n = n\log m log m n = n log m Differentiating both sides w.r.t. x x x
1 y d y d x = x ⋅ sec 2 x tan x + log ( tan x ) \frac1y\,\frac{dy}{dx} = x\cdot\frac{\sec^2x}{\tan x} + \log(\tan x) y 1 d x d y = x ⋅ tan x sec 2 x + log ( tan x ) Why?
Left side: y depends on x x x d d x log y = 1 y ⋅ d y d x \frac{d}{dx}\log y = \frac1y\cdot\frac{dy}{dx} d x d log y = y 1 ⋅ d x d y
Right side: the product rule ( u v ) ′ = u v ′ + v u ′ (uv)' = uv' + vu' ( uv ) ′ = u v ′ + v u ′ u = x u = x u = x v = log ( tan x ) v = \log(\tan x) v = log ( tan x )
d d x log ( tan x ) = sec 2 x tan x \frac{d}{dx}\log(\tan x) = \frac{\sec^2x}{\tan x} d x d log ( tan x ) = tan x sec 2 x and d d x x = 1 \frac{d}{dx}x = 1 d x d x = 1
u = x , u ′ = 1 v = log ( tan x ) , v ′ = 1 tan x ⋅ sec 2 x \begin{aligned} u &= x, & u' &= 1 \\ v &= \log(\tan x), & v' &= \tfrac{1}{\tan x}\cdot\sec^2x \end{aligned} u v = x , = log ( tan x ) , u ′ v ′ = 1 = t a n x 1 ⋅ sec 2 x u v ′ + v u ′ = x ⋅ sec 2 x tan x + log ( tan x ) ⋅ 1 \begin{aligned} uv' + vu' &= x\cdot\frac{\sec^2x}{\tan x} \\ &\quad + \log(\tan x)\cdot 1 \end{aligned} u v ′ + v u ′ = x ⋅ tan x sec 2 x + log ( tan x ) ⋅ 1 d y d x = y [ x sec 2 x tan x + log ( tan x ) ] \begin{aligned} \frac{dy}{dx} &= y\Big[\frac{x\sec^2x}{\tan x} \\ &\qquad + \log(\tan x)\Big] \end{aligned} d x d y = y [ tan x x sec 2 x + log ( tan x ) ] Why?
Multiply both sides by y. Then put back y = ( tan x ) x y = (\tan x)^x y = ( tan x ) x x x x
∴ d y d x = ( tan x ) x [ x sec 2 x tan x \therefore\ \dfrac{dy}{dx} = (\tan x)^x\Big[\dfrac{x\sec^2x}{\tan x} ∴ d x d y = ( tan x ) x [ tan x x sec 2 x + log ( tan x ) ] + \log(\tan x)\Big] + log ( tan x ) ]
Where this comes from Chapter 5, Continuity and Differentiability : Implicit and logarithmic differentiation.
Question type: (a) Implicit ; (b) Logarithmic differentiation
How to spot it: click to see (a) Type 5. Implicit: x x x y y y
Spot it: an equation such as x 2 + x y + y 2 = 100 x^2 + xy + y^2 = 100 x 2 + x y + y 2 = 100 y y y
Here: x x x y y y x = sin θ x = \sin\theta x = sin θ y = sin ϕ y = \sin\phi y = sin ϕ sin − 1 x − sin − 1 y = 2 cot − 1 a \sin^{-1}x - \sin^{-1}y = 2\cot^{-1}a sin − 1 x − sin − 1 y = 2 cot − 1 a y y y d y d x \frac{dy}{dx} d x d y
(b) Type 7. Logarithmic differentiation: x x x
Spot it: a power whose exponent contains x x x x x x^x x x x sin x x^{\sin x} x s i n x ( log x ) cos x (\log x)^{\cos x} ( log x ) c o s x
Here: x x x tan x \tan x tan x log y = x log ( tan x ) \log y = x\log(\tan x) log y = x log ( tan x ) y = ( tan x ) x y = (\tan x)^x y = ( tan x ) x
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