Continuity and Differentiability Test

Result

Continuity and Differentiability Test - 9

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Question 1
1 / -0.25

The differential coefficient dy/dx of the function y^x = x^y

A

B

C

D

Solution

d(x^y )=d(e^{(y ⋅log(x))} )
=e^(y ⋅log(x))d(y ⋅log(x))
=(x^y )(dy ⋅log(x) + y ⋅d(log(x))
=(x^y )
d(y^x ) = (y^x )(log(y)dx + x/ydy).
Since d(x^y ) = d(y^x ) , and simplifying by x^y = y^x , we get
log(y)dx + x/ydy = log(x)dy + y/xdx.
Removing the denominators leads to:
xylog(y)dx + x² dy = xylog(x)dy + y² dx
(xylog(y) −y² )dx = (xylog(x) −x² )dy
dy/dx = (xylog(y)−y² )/(xy ⋅log(x)−x² )
Question 2
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The derivative of 2^x tan x is

A
2^x log 2 [ sec² x + tan x]

B
2^x tan x [sec x + log 2]

C
2^x [sec² x + log 2 tan x]

D
2^x [sec² x + tan x]

Solution

dy/dx = 2^x (tanx)'+ tanx (2^x )'
dy/dx = 2^x (secx)² + 2^x tanx log2 , {since , (a^x )'= a^x loga (x)'}
dy/dx = 2^x [(Secx)² + log2 tanx]
Question 3
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The differential coefficient of (log x)^tanx is:

A

B

C

D

Solution

y = (lnx)^tanx
ln(y) = tanx(ln(lnx))
d(lny)/dx = d(tanx(ln(lnx)))/dx
Using product rule -
(1/y)dy/dx = (secx)² (ln(lnx)) + (1 ÷xlnx)tanx
dy/dx = [(secx)² (ln(lnx)) + (1 ÷xlnx)tanx ]×y
dy/dx = [(secx)² (ln(lnx)) + (1 ÷xlnx)tanx ]×[(lnx)^tanx ]
Question 4
1 / -0.25

The differential coefficient of the function f(x) = a^{sin x} , where a is positive constant is:

A
cos x. log sin x

B
a^{sin x} . cos x. log a

C
a^{sin x} log sin x

D
a^{sin x} . cos x. log sin x
Question 5
1 / -0.25

A
0

B
1

C
1/2

D
none of these

Solution
Question 6
1 / -0.25

A

B

C

D

Solution

y = (1 - log x)/(1 + log x)
Applying the Quoitent rule and differential for ln x
dy / dx = [(1 - ln x)(1 / x) - (1+ ln x)( -1 / x)]/(1 - ln x)²
= [(1 - ln x + 1 + ln x)]/x(1 + ln x)²
= 2 / x(1+ln x)²
Question 7
1 / -0.25

If x = a cos φand y = b sin φthen dy/dx =

A

B

C

D
Question 8
1 / -0.25

If sin(x + y) = log(x + y), :

A
1

B
2

C
-1

D
-2

Solution

Given equation,
Question 9
1 / -0.25

The differential coffcient of the equation y^x = e^{(x - y)} is :

A

B

C

D

Solution

Taking log both the sides
xlogy = (x-y)loge
Differentiate it with respect to x, we get
x/y dy/dx = logy = loge - xloge dy/dx
dy/dx = (loge - logy)/(x/y + loge)
= y(1 - loge)/(x + y)
Question 10
1 / -0.25

A

B

C

D

Solution

On differentiating both sides with respect to x