Test: Differentiation by Taking Log (20 August) - JEE MCQ

• log mⁿ = n log m
• Differentiation by part: 
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Calculation :
Given that y = (x^x)^x,
Taking log on both sides
log y = log (x^x)^x
Using the formula (1)
⇒ log y = x log x^x
⇒ log y = x² log x
Diff. with respect to x 

Doing diff. by part as discussed in formula (2)

Put x = 1 in above equation

Concept:
Product rule: (f.g)'(x) = f '(x).g(x) + f(x).g'(x)
Formula used :

Calculation:
Given,  y = x√x,
Taking log on both sides,
ln y = √x ln x
Differentiating both sides,

∴ The correct option is (4).

Calculation:
Given that,
f(x) = log_e [log_e x]

Put x = e in the above equation then

The correct answer is option "1".

Concept:
Product rule: (f.g)'(x) = f '(x).g(x) + f(x).g'(x)
Formula used :

Calculation:
Given,  y = x√x,
Taking log on both sides,
ln y = √�� ln x
Differentiating both sides,

∴ The correct option is (4).

Formula used:

1. 
2. log m + log n = log mn
3. log mⁿ = n log m  

Calculation:
Given that, 
Doing integration on both sides

⇒ ln y = x ln 5 + ln C
Using the logarithmic property (2) & (3)
⇒ ln y = ln 5^x + ln C
⇒ ln y = ln C5^x
⇒ y = C ×  5^x          ------(1)
According to the question, y(0) = ln 5
ln 5 = C × 5⁰ 
⇒ C = ln 5
Put this value in equation (1)
⇒ y = 5^x ln 5 
Put x = 1
∴ y(1) = 5 ln 5

Concept:
Suppose that we have two functions f(x) and g(x) and they are both differentiable.

Calculation:
y = x^x
Taking log both sides, we get
⇒ log y = log x^x   (∵ log mⁿ = n log m)
⇒ log y = x log x
Differentiating with respect to x, we get

put x = 1

Concept:

Calculation:
Let y = 
Taking log both sides, we get
ln y = ln 
ln y = e^x (ln x) (∵ log mⁿ = n log m)
Differentiating with respect to x, we get

Concept:
log ( a + b ) = log a + log b

Calculation:
 x^m yⁿ  = 2(x + y)^{m + n} 
Taking log on both sides , we get 
⇒ log(x^m yⁿ) = log[2(x + y)^{m + n}]
⇒ m log x + n log y = log 2 + (m + n) log (x + y)
on differentiating both sides with respect to x , we get

Concept:

Product rule: 

Calculation:
Let, y = x^2x 
Taking log on both the sides, we get 
log y = log( x^2x ) = 2x logx

Now, taking derivatives, 

(∵ y = x^2x )
Hence, option (4) is correct.

Concept:

Calculation:
y = x^ln x 
Taking log both sides, we ge
ln y = ln x^ln x
ln y = (ln x)²   (∵ log mⁿ = n log m)
Differentiating with respect to x, we get