# Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.2 | Set 1

• Last Updated : 30 Apr, 2021

### (i) 2/x

Solution:

Given that f(x) = 2/x

By using the formula

f'(x) =

We get

### (ii) 1/√x

Solution:

Given that f(x) = 1/√x

By using the formula

We get

### (iii) 1/x3

Solution:

We have f(x) = 1/x3

By using the formula

We get

### (iv) (x2 + 1)/x

Solution:

Given that f(x) = (x2 + 1)/x

By using the formula

We get

### (v) (x2 – 1)/x

Solution:

Given that f(x) = (x2 – 1)/x

By using the formula

We get

### (vi) (x + 1)/(x + 2)

Solution:

Given that f(x) = (x + 1)/(x + 2)

By using the formula

We get

= 1/(x + 2)2

### (vii) (x + 2)/(3x + 5)

Solution:

Given that f(x) = (x + 2)/(3x + 5)

By using the formula

We get

### (viii) kxn

Solution:

Given that f(x) = kxn

By using the formula

We get

= k nxn-1+ 0 + 0 …

= k nxn-1

### (ix) 1/√(3 – x)

Solution:

Given that f(x) = 1/√(3-x)

By using the formula

We get

### (x) x2 + x + 3

Solution:

Given that f(x) = x2 + x + 3

By using the formula

We get

= 2x + 0 + 1

= 2x + 1

### (xi) (x + 2)3

Solution:

Given that f(x) = (x + 2)3

By using the formula

We get

= 3(x + 2)2

### (xii) x3 + 4x2 + 3x + 2

Solution:

Given that f(x) = x3 + 4x2 + 3x + 2

By using the formula

We get

= 3x2 + 8x + 3

### (xiii) (x2 + 1)(x – 5)

Solution:

Given that f(x) = (x2+1)(x-5)

By using the formula

We get

= 3x2 – 10x + 1

### (xiv) √(2x2 + 1)

Solution:

Given that f(x) = √(2x2 + 1)

By using the formula

We get

On multiplying numerator and denominator by

We get

### (xv) (2x + 3)/(x – 2)

Solution:

Given that f(x) = (2x + 3)/(x – 2)

By using the formula

We get

### (i) e-x

Solution:

Given that f(x) = e-x

By using the formula

We get

= -e-x

### (ii) e3x

Solution:

Given that f(x) = e3x

By using the formula

We get

Multiplying numerator and denominator by 3.

Here,

= 3e3x

### (iii) eax+b

Solution:

Given that f(x) = eax+b

By using the formula

We get

On multiplying numerator and denominator by a

Since

= aeax+b

### (iv) xex

Solution:

Given that f(x) = xex

By using the formula

We get

= xex + ex

= ex(x + 1)

### (v) x2 ex

Solution:

Given that f(x) = x2ex

By using the formula

We get

= x2ex + ex(0 + 2x)

= x2ex + 2xex

= ex(x2 + 2x)

### (vi)

Given that f(x) =

By using the formula

We get

### (vii) e√(2x)

Solution:

Given that f(x) =

By using the formula

We get

On multiplying numerator and denominator by

we get

Again multiplying numerator and denominator by

we get

### (viii) e√(ax + b)

Solution:

Given that f(x) = e√(ax+b)

By using the formula

We get

On multiplying numerator and denominator by

we get

Again multiplying numerator and denominator by

we get

=

### (ix) a√x

Solution:

Given that f(x) = a√x = e√xloga

By using the formula

We get

On multiplying numerator and denominator by

we get

f”(x) =

On multiplying numerator and denominator by

we get

f'(x) =

logea

### (x)

Solution:

Given that f(x) =

By using the formula

We get

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