Engineering Mathematics Multiple Choice Questions on “Implicit Differentiation”.
1. Find the differentiation of x3 + y3 – 3xy + y2 = 0?
a) (frac{(x^2-y)}{x-y^2-2y})
b) (frac{(3x^2-3y)}{3x-3y^2-2y})
c) (frac{(3x^3-3y)}{3x-3y^2-2y})
d) (frac{(3x^2-y)}{3x-3y^2-y})
Answer: b
Explanation: Differentiation of x3 is 3x2
differentiation of y3 is 3y2 (frac{dy}{dx})
differentiation of -3xy is [-3y -3x (frac{dy}{dx})]
differentiation of y2 is 2y (frac{dy}{dx})
Hence,
(frac{d(x^3+y^3-3xy+y^2)}{dx}=0)
(3x^2+3y^2 frac{dy}{dx}-3y-3x frac{dy}{dx}+2y frac{dy}{dx} = 0)
(frac{dy}{dx}=frac{(3x^2-3y)}{3x-3y^2-2y})
2. x3 Sin(y) + Cos(x) y3 = 0, its differentiation is?
a) (-frac{[x^3 Sin(y)-3y^2 Sin(x)]}{[x^2 Cos(y)+y^3 Cos(x)]})
b) (-frac{[3x^2 Sin(y)-y^3 Sin(x)]}{[x^3 Cos(y)+3y^2 Cos(x)]})
c) (-frac{[3x^3 Sin(y)-y^3 Sin(x)]}{[x^3 Cos(y)+3y^3 Cos(x)]})
d) 0
Answer: b
Explanation:
(frac{d}{dx}[x^3 Sin(y)+Cos(x) y^3])= 0
(3x^2 Sin(y)+x^3 Cos(y) frac{dy}{dx}-Sin(x) y^3+3y^2 frac{dy}{dx} Cos(x)=0)
(frac{dy}{dx}=-frac{[3x^2 Sin(y)-y^3 Sin(x)]}{[x^3 Cos(y)+3y^2 Cos(x)]})
3. Find the differentiation of x4 + y4 = 0.
a) – x3⁄y4
b) – x4⁄y3
c) – x3⁄y3
d) x3⁄y3
Answer: c
Explanation: x4 + y4 = 0
4x3 + 4y3dy⁄dx = 0
dy⁄dx = – x3⁄y3
dy⁄dx = Sec2 (x)Sec(x) ex + Sec2 (x)Tan(x) ex + ex Tan(x)Sec(x)
dy⁄dx = Sec2 (x) ex [Sec(x)+Tan(x)] + ex Tan(x)Sec(x)
4. Find differentiation of xSin(x) + ayCos(x) + Tan(y) = 0.
a) (frac{[ayCos(x)-Sin(x)+Cos(x)]}{[aCos(x)+Sec^2 (y)]})
b) (frac{[ayCos(x)-Sin(x)+xCos(x)]}{[Cos(x)+Sec^2 (y)]})
c) (frac{[ayCos(x)-Sin(x)+xCos(x)]}{[aCos(x)+Sec^2 (y)]})
d) (frac{[ayCos(x)-Cos(x)+xCos(x)]}{[aCos(x)+Sec^2 (y)]})
Answer: c
Explanation: xSin(x) + ayCos(x) + Tan(y) =0
Differentiation of above eqn. is
(Sin(x) + xCos(x) – ayCos(x) + aCos(x) frac{dy}{dx} + Sec^2 (y) frac{dy}{dx}=0)
(frac{dy}{dx}=frac{[ayCos(x)-Sin(x)+xCos(x)]}{[aCos(x)+Sec^2 (y)]})
5. Find the derivative of Tan(x) = Tan(y).
a) (frac{1+x^2}{1+y^2})
b) (frac{1+y}{1+x^2})
c) (frac{1+y^2}{1+x^2})
d) (frac{1+y^2}{1+x})
Answer: c
Explanation:
Tan(y)=Tan(x)
(frac{1}{1+y^2} frac{dy}{dx}=frac{1}{1+x^2})
(frac{dy}{dx}=frac{1+y^2}{1+x^2})
6. Implicit functions are those functions ____________
a) Which can be solved for a single variable
b) Which can not be solved for a single variable
c) Which can be eliminated to give zero
d) Which are rational in nature.
Answer: b
Explanation: Implicit functions are those functions, Which can not be solved for a single variable.
For ex, f(x,y) = x3 +y3-3xy = 0.
7. Evaluate y44 + 3xy3 + 6x2 y2 – 7y + 8 = 0.
a) (frac{(7-12x^2 y-9xy^4-4y^3)}{(3y^3+12xy^2)})
b) (frac{(7-12x^2 y-9xy^2-4y^3)}{(3y^3+12xy^2)})
c) (frac{(7-12x^2 y-9xy^2-4y^3)}{(3y^4+12xy^2)})
d) (frac{(7-12x^4 y-9xy^2-4y^3)}{(3y^3+12xy^2)})
Answer: b
Explanation: y44 + 3xy3 + 6x2 y2 – 7y + 8 = 0.
Differentiating it we get
(4y^3 frac{dy}{dx}+3[y^3+3xy^2 frac{dy}{dx}]+6[2xy^2+2x^2 y frac{dy}{dx}]-7 frac{dy}{dx})=0
(frac{dy}{dx}=frac{(7-12x^2 y-9xy^2-4y^3 )}{(3y^3+12xy^2)})
8. If Sin(y)=Sin(-1) (y) then?
a) (1-y2)(1 – Cos2 y) = 1
b) (1-y2)(1 – Sin2 y) = 1
c) (1-y2)(1 – Siny)=1
d) (1-y2)(1 – Cosy)=1
Answer: b
Explanation: Sin(y)=Sin(-1) (y)
Differentiating both sides
(Cos(y) frac{dy}{dx}=frac{1}{sqrt{1-y^2}} frac{dy}{dx})
(1-y2)(1-Sin2 y)=1
9. If Cos(y)=Cos(-1) (y) then?
a) (1 – y2)(1 – Cos2 (y))=1
b) (1 – y2)(1 – Cos(y))=1
c) (1 – y2)(1 – Sin2 (y))=1
d) (1 – y2)(1 – Sin(y))=1
Answer: a
Explanation: Cos(y)=Cos(-1) (y)
Differentiating both sides
-Sin(y) = (-frac{1}{sqrt{1-y^2}})
(1 – y2)(1 – Cos2 (y)) = 1.
10. If y2 + xy + x2 – 2x = 0 then d2y⁄dx2 =?
a) ((2y+x) frac{d^2 y}{dx^2}+(frac{dy}{dx})^2+2 frac{dy}{dx}+2=0)
b) ((2y+x) frac{d^2 y}{dx^2}+2(frac{dy}{dx})^2+frac{dy}{dx}+2=0)
c) ((2y+x) frac{d^2 y}{dx^2}+2(frac{dy}{dx})^2+2 frac{dy}{dx}+2=0)
d) (x frac{d^2 y}{dx^2}+2(frac{dy}{dx})^2+2 frac{dy}{dx}+2=0)
Answer: c
Explanation:
(y^2+xy+x^2-2x=0)
(2y frac{dy}{dx}+x frac{dy}{dx}+y+2x-2=0)
(frac{2yd^2 y}{dx^2}+2(frac{dy}{dx})^2+x (frac{d^2 y}{dx^2})+frac{dy}{dx}+frac{dy}{dx}+2=0)
((2y+x) frac{d^2 y}{dx^2}+2(frac{dy}{dx})^2+2 frac{dy}{dx}+2=0)
11. If the velocity of car at time t(sec) is directly proportional to the square of its velocity at time (t-1)(sec). Then find the ratio of acceleration at t=10sec to 9sec if proportionality constant is k=10 sec/mt and velocity at t=9sec is 10 mt/sec.
a) 100
b) 200
c) 150
d) 250