BITSAT 2021
Exam BITSAT
Year 2021
Subject Mathematics
Chapter Differentiation BITSAT 2021 · Mathematics · Differentiation
If x= 1-t^ 2 1+t^ 2 and y= 2 t 1+t^ 2 , then (d y)/(d x) is equal...
If
x = 1 − t 2 1 + t 2 x=\frac{1-t^{2}}{1+t^{2}} x = 1 + t 2 1 − t 2 and
y = 2 t 1 + t 2 y=\frac{2 t}{1+t^{2}} y = 1 + t 2 2 t , then
d y d x \frac{d y}{d x} d x d y is equal to :
Options A − y x -\frac{y}{x} − x y B y x \frac{y}{x} x y C − x y -\frac{x}{y} − y x D x y \frac{x}{y} y x Correct answer C. − x y -\frac{x}{y} − y x
Step-by-step solution
Let
x = 1 − t 2 1 + t 2 x=\frac{1-t^{2}}{1+t^{2}} x = 1 + t 2 1 − t 2 and
y = 2 t 1 + t 2 y=\frac{2 t}{1+t^{2}} y = 1 + t 2 2 t Put
t = tan θ t=\tan \theta t = tan θ , we get
x = 1 − tan 2 θ 1 + tan 2 θ x=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} x = 1 + t a n 2 θ 1 − t a n 2 θ and
y = 2 tan θ 1 + tan 2 θ ⇒ x = cos 2 θ y=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\Rightarrow x=\cos 2 \theta y = 1 + t a n 2 θ 2 t a n θ ⇒ x = cos 2 θ and
y = sin 2 θ ∴ d x d θ = − 2 sin 2 θ y=\sin 2 \theta\therefore \frac{\mathrm{dx}}{\mathrm{d} \theta}=-2 \sin 2 \theta y = sin 2 θ ∴ d θ dx = − 2 sin 2 θ and
d y d θ = 2 cos 2 θ \frac{\mathrm{dy}}{\mathrm{d} \theta}=2 \cos 2 \theta d θ dy = 2 cos 2 θ Now,
d y d x = d y d θ × d θ d x = − cos 2 θ sin 2 θ = − x y \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{dx}}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y} dx dy = d θ dy × dx d θ = − s i n 2 θ c o s 2 θ = − y x
Practice Differentiation on Quantrex Academy →
← Previous question If 2^ x +2^ y =2^ x+y , then (d y)/(d x)=Next question → (d)/(d x) ( ^ -1 (1+ (x)/(2))/(1- (x)/(2)) ) is equal to
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