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

### Question 1. Differentiate each of the following using first principles:

### (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/x^{3}

**Solution:**

We have f(x) = 1/x

^{3}By using the formula

We get

=

=

=

=

=

=

### (iv) (x^{2 }+ 1)/x

**Solution:**

Given that f(x) = (x

^{2 }+ 1)/xBy using the formula

We get

=

=

=

=

=

=

### (v) (x^{2 }– 1)/x

**Solution:**

Given that f(x)

=(x^{2 }– 1)/xBy 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) kx^{n}

**Solution:**

Given that f(x) = kx

^{n}By using the formula

We get

=

=

=

= k nx

^{n-1}+ 0 + 0 …= k nx

^{n-1}

### (ix) 1/√(3 – x)

**Solution:**

Given that f(x) = 1/√(3-x)

By using the formula

We get

=

=

=

=

=

=

=

### (x) x^{2 }+ x + 3

**Solution:**

Given that f(x) = x

^{2 }+ x + 3By 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) x^{3 }+ 4x^{2} + 3x + 2

**Solution:**

Given that f(x) = x

^{3 }+ 4x^{2}+ 3x + 2By using the formula

We get

=

=

=

=

= 3x

^{2 }+ 8x + 3

### (xiii) (x^{2 }+ 1)(x – 5)

**Solution:**

Given that f(x) = (x

^{2}+1)(x-5)By using the formula

We get

=

=

=

=

= 3x

^{2 }– 10x + 1

### (xiv) √(2x^{2 }+ 1)

**Solution:**

Given that f(x) = √(2x

^{2 }+ 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

=

=

=

=

### Question 2. Differentiate each of the following using first principles:

### (i) e^{-x}

**Solution:**

Given that f(x) = e

^{-x}By using the formula

We get

=

=

=

= -e

^{-x}

### (ii) e^{3x}

**Solution:**

Given that f(x) = e

^{3x}By using the formula

We get

=

=

=

Multiplying numerator and denominator by 3.

=

Here,

= 3e

^{3x}

### (iii) e^{ax+b}

**Solution:**

Given that f(x) = e

^{ax+b}By using the formula

We get

=

=

=

=

On multiplying numerator and denominator by a

Since

= ae

^{ax+b}

### (iv) xe^{x}

**Solution:**

Given that f(x) = xe

^{x}By using the formula

We get

=

=

=

= xe

^{x }+ e^{x}= e

^{x}(x + 1)

### (v) x^{2 }e^{x}

**Solution:**

Given that f(x) = x

^{2}e^{x}By using the formula

We get

=

=

= x

^{2}e^{x}+ e^{x}(0 + 2x)= x

^{2}e^{x}+ 2xe^{x}= e

^{x}(x^{2}+ 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) =

=

=

= log

_{e}a

### (x)

**Solution:**

Given that f(x) =

By using the formula

We get

=

=

=

=

=

=

=

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