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QUESTION: 1

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

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 + x^{2}dy = xylog(x)dy + y^{2}dx

(xylog(y) − y^{2})dx = (xylog(x) − x^{2})dy

dy/dx = (xylog(y)−y^{2})/(xy⋅log(x)−x^{2})

QUESTION: 2

The derivative of 2^{x} tan x is

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

The differential coefficient of (log x)^{tanx} is:

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

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

Solution:

QUESTION: 5

Solution:

QUESTION: 6

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

If x = a cos φ and y = b sin φ then dy/dx =

Solution:

QUESTION: 8

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

Solution:

Given equation,

QUESTION: 9

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

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

Solution:

On differentiating both sides with respect to x

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