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
View Answer

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

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) \(\frac{(3x^2-y)}{3x-3y^2-y}\)
View Answer

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

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

Answer: c
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)
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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) \(\frac{[ayCos(x)-Cos(x)+xCos(x)]}{[aCos(x)+Sec^2 (y)]}\)
View Answer

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) \(\frac{1+y^2}{1+x}\)
View Answer

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

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

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
\(\frac{dy}{dx}=\frac{(7-12x^2 y-9xy^2-4y^3 )}{(3y^3+12xy^2)}\)
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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
View Answer

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
View Answer

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.
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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) \(x \frac{d^2 y}{dx^2}+2(\frac{dy}{dx})^2+2 \frac{dy}{dx}+2=0\)
View Answer

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
View Answer

Answer: b
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)
View Answer

Answer: c
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
View Answer

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

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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