CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Continuity & Differentiability

5.2. Derivatives of Implicit Functions.

Q.1. If y = √x + 1/√x,
show that 2x dy/dx + y = 2√x.

Solution :

We have, y = √x + 1/√x
Therefore, dy/dx = d/dx(√x) + d/dx(1/√x)
= 1/(2√x) – 1/(2x√x)
Or, 2x dy/dx = √x – 1/√x
Or, 2x dy/dx + y = √x – 1/√x + y
=√x – 1/√x + √x + 1/√x
= 2√x. [Proved.]

Q.2. If xy + y² = tan x + y, find dy/dx.

Solution :

We have, xy + y² = tan x + y,
Differentiating w.r.t., x we get
x dy/dx + y.1 + 2y dy/dx = sec² x + dy/dx
Or, x dy/dx + 2y dy/dx – dy/dx = – y + sec² x
Or, dy/dx [x + 2y – 1] = – y + sec² x
Therefore, dy/dx = (– y + sec² x)/(x + 2y – 1). [Ans.]

Q.3. If x√(1 + y) + y√(1 + x) = 0, for – 1 < x, y < 1, prove the following :
dy/dx = – 1/(1 + x)².

Solution :

We have, x√(1 + y) + y√(1 + x) = 0
Or, x√(1 + y) = – y√(1 + x)
Squaring both sides, we get
x²(1 + y) = y² (1 + x)
Or, x² + x²y – y² – xy² = 0
Or, x² – y² + x²y – xy² = 0
Or, (x + y)(x – y) + x y (x – y) = 0
Or, (x – y)(x + y + x y) = 0
Either, x – y = 0 Or, x + y + x y = 0
But x ≠ y , hence, x + y + x y = 0
Or, y(1 + x) = – x
Or, y = – x/(1 + x)
Therefore, dy/dx = – [{(1 + x).1 – x.(0 + 1)}/(1 + x)²]
= – (1 + x – x)/(1 + x)²
= – 1/(1 + x)². [Ans.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289