# Engineering Mathematics Questions and Answers – Implicit Differentiation

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Implicit Differentiation”.

1. 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

Explanation:
$$\frac{d}{dx}[x^3 Sin(y)+Cos(x) y^3]$$= 0
$$3x^2 Sin(y)+x^3 Cos(y) \frac{dy}{dx}-Sin(x) y^3+3y^2 \frac{dy}{dx} Cos(x)=0$$
$$\frac{dy}{dx}=- \frac{[3x^2 Sin(y)-y^3 Sin(x)]}{[x^3 Cos(y)+3y^2 Cos(x)]}$$

2. 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) $$\frac{(3x^2-y)}{3x-3y^2-y}$$

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,
$$\frac{d(x^3+y^3-3xy+y^2)}{dx}=0$$
$$3x^2+3y^2 \frac{dy}{dx}-3y-3x \frac{dy}{dx}+2y \frac{dy}{dx} = 0$$
$$\frac{dy}{dx}=\frac{(3x^2-3y)}{3x-3y^2-2y}$$

3. Find the differentiation of x4 + y4 = 0.
a) – x3y4
b) – x4y3
c) – x3y3
d) x3y3

Explanation: x4 + y4 = 0
4x3 + 4y3 dydx = 0
dydx = – x3y3
dydx = Sec2 (x)Sec(x) ex + Sec2 (x)Tan(x) ex + ex Tan(x)Sec(x)
dydx = 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) $$\frac{[ayCos(x)-Cos(x)+xCos(x)]}{[aCos(x)+Sec^2 (y)]}$$

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) $$\frac{1+y^2}{1+x}$$

Explanation:
Tan(y)=Tan(x)
$$\frac{1}{1+y^2} \frac{dy}{dx}=\frac{1}{1+x^2}$$
$$\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$$
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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.

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) $$\frac{(7-12x^4 y-9xy^2-4y^3)}{(3y^3+12xy^2)}$$

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

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

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 d2ydx2 =?
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) $$x \frac{d^2 y}{dx^2}+2(\frac{dy}{dx})^2+2 \frac{dy}{dx}+2=0$$

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

Explanation:
Given,v(t)=kv2 (t-1)
Differentiating w.r.t time we get
dv(t)dt = 2kv(t-1) dv(t – 1)dt
a(t) = 2*10*10 a(t-1)
a(t)a(t – 1) = 200.

12. If z(x,y) = 2Sin(x)+Cos(y)Sin(x) find d2z(xy)dxdy= ?
a) –Cos(y)Cos(x)
b) -Sin(y)Sin(x)
c) –Sin(y)Cos(x)
d) -Cos(y)Sin(x)

Explanation: z(x,y) = 2Sin(x) + Cos(y)Sin(x)
Hence,
$$\frac{d^2 z(x,y)}{dxdy}=\frac{d}{dx} \frac{dz(x,y)}{dy}=\frac{d}{dx}[\frac{d}{dy} (2Sin(x)+Cos(y)Sin(x))]$$
=$$-\frac{d}{dx}$$ Sin(y)Sin(x)= -Sin(y)Cos(x)

13. If the car is having a displace from point 1 to point 2 in t sec which is given by equation y(x) = x2 + x + 1. Then?
a) Car is moving with constant acceleration
b) Car is moving with constant velocity
c) Neither acceleration nor velocity is constant
d) Both acceleration and velocity is constant

Explanation: y(x) = x2 + x + 1
Velocity is, v = dydx = 2x + 1 (not constant)
Acceleration is a = dydx = 2 (constant).

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