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RD Sharma Class 12 Solutions - Differentiation - 2 | Mathematics (Maths) Class 12 - JEE PDF Download

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27. Question
Find , when
y = (tan x)
cot x
 + (cot x)
tan x
Answer
let y = (tan x)
cot x
 + (cot x)
tan x
? y = a + b
where a= (tan x)
cot x
 ; b = (cot x)
tan x
a= (tan x)
cot x
Taking log both the sides:
? log a= log (tan x)
cot x
? log a= cot x log (tan x)
{log x
a
 = alog x}
Differentiating with respect to x:
b = (cot x)
tan x
Page 2


27. Question
Find , when
y = (tan x)
cot x
 + (cot x)
tan x
Answer
let y = (tan x)
cot x
 + (cot x)
tan x
? y = a + b
where a= (tan x)
cot x
 ; b = (cot x)
tan x
a= (tan x)
cot x
Taking log both the sides:
? log a= log (tan x)
cot x
? log a= cot x log (tan x)
{log x
a
 = alog x}
Differentiating with respect to x:
b = (cot x)
tan x
Taking log both the sides:
? log b= log (cot x)
tan x
? log b= tan x log (cot x)
{log x
a
 = alog x}
Differentiating with respect to x:
28. Question
Find , when
Answer
? y = a + b
Page 3


27. Question
Find , when
y = (tan x)
cot x
 + (cot x)
tan x
Answer
let y = (tan x)
cot x
 + (cot x)
tan x
? y = a + b
where a= (tan x)
cot x
 ; b = (cot x)
tan x
a= (tan x)
cot x
Taking log both the sides:
? log a= log (tan x)
cot x
? log a= cot x log (tan x)
{log x
a
 = alog x}
Differentiating with respect to x:
b = (cot x)
tan x
Taking log both the sides:
? log b= log (cot x)
tan x
? log b= tan x log (cot x)
{log x
a
 = alog x}
Differentiating with respect to x:
28. Question
Find , when
Answer
? y = a + b
a = (sin x)
x
Taking log both the sides:
? log a= log (sin x)
x
? log a= x log (sin x)
{log x
a
 = alog x}
Differentiating with respect to x:
Put the value of a= (sin x)
x
 :
Differentiating with respect to x:
Page 4


27. Question
Find , when
y = (tan x)
cot x
 + (cot x)
tan x
Answer
let y = (tan x)
cot x
 + (cot x)
tan x
? y = a + b
where a= (tan x)
cot x
 ; b = (cot x)
tan x
a= (tan x)
cot x
Taking log both the sides:
? log a= log (tan x)
cot x
? log a= cot x log (tan x)
{log x
a
 = alog x}
Differentiating with respect to x:
b = (cot x)
tan x
Taking log both the sides:
? log b= log (cot x)
tan x
? log b= tan x log (cot x)
{log x
a
 = alog x}
Differentiating with respect to x:
28. Question
Find , when
Answer
? y = a + b
a = (sin x)
x
Taking log both the sides:
? log a= log (sin x)
x
? log a= x log (sin x)
{log x
a
 = alog x}
Differentiating with respect to x:
Put the value of a= (sin x)
x
 :
Differentiating with respect to x:
29 A. Question
Find , when
y = x
cos x
 + (sin x)
tan x
Answer
let y = x
cos x
 + (sin x)
tan x
? y = a + b
where a= x
cos x
 ; b = (sin x)
tan x
a= x
cos x
Taking log both the sides:
? log a= log (x)
cos x
? log a= cos x log x
{log x
a
 = alog x}
Differentiating with respect to x:
Page 5


27. Question
Find , when
y = (tan x)
cot x
 + (cot x)
tan x
Answer
let y = (tan x)
cot x
 + (cot x)
tan x
? y = a + b
where a= (tan x)
cot x
 ; b = (cot x)
tan x
a= (tan x)
cot x
Taking log both the sides:
? log a= log (tan x)
cot x
? log a= cot x log (tan x)
{log x
a
 = alog x}
Differentiating with respect to x:
b = (cot x)
tan x
Taking log both the sides:
? log b= log (cot x)
tan x
? log b= tan x log (cot x)
{log x
a
 = alog x}
Differentiating with respect to x:
28. Question
Find , when
Answer
? y = a + b
a = (sin x)
x
Taking log both the sides:
? log a= log (sin x)
x
? log a= x log (sin x)
{log x
a
 = alog x}
Differentiating with respect to x:
Put the value of a= (sin x)
x
 :
Differentiating with respect to x:
29 A. Question
Find , when
y = x
cos x
 + (sin x)
tan x
Answer
let y = x
cos x
 + (sin x)
tan x
? y = a + b
where a= x
cos x
 ; b = (sin x)
tan x
a= x
cos x
Taking log both the sides:
? log a= log (x)
cos x
? log a= cos x log x
{log x
a
 = alog x}
Differentiating with respect to x:
b = (sin x)
tan x
Taking log both the sides:
? log b= log (sin x)
tan x
? log b= tan x log (sin x)
{log x
a
 = alog x}
Differentiating with respect to x:
29 B. Question
Find , when
y = x
x
 + (sin x)
x
Answer
let y = x 
x
 + (sin x) 
x
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RD Sharma Class 12 Solutions - Differentiation - 2 | Mathematics (Maths) Class 12 - JEE

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