### Find the second order derivatives of the functions given in Exercises 1 to 10.

### Question 1. x^{2}+ 3x + 2

**Solution:**

Here, y = x

^{2}+ 3x + 2

First derivative,= 2x+ 3

Second derivative,=

= 2

### Question 2. x^{20}

**Solution:**

Here, y = x

^{20}

First derivative,= 20x

^{20-1}= 20x

^{19}

Second derivative,=

= 20(19x

^{19-1})

= 380x^{18}

### Question 3. x . cos x

**Solution:**

Here, y = x . cos x

First derivative,Using product rule

= x + cos x

= x (-sin x)+ cos x (1)

= – x sin x+ cos x

Second derivative,=

=

Using product rule,

= -x + sin x + (- sin x)

= -x (cos x) + sin x (-1) – sin x

= – ( x cos x + 2 sin x)

### Question 4. log x

**Solution:**

Here, y = log x

First derivative,= 1/x

Second derivative,=

Using division rule,

=

=

=

### Question 5. x^{3 }log x

**Solution:**

Here, y = x

^{3}. log x

First derivative,Using product rule

= x

^{3}+ log x= x

^{3}() + log x (3x^{2})= x

^{2}+ 3x^{2}log x

Second derivative,=

= +

Using product rule,

= 2x + 3 (x

^{2}– log x)= 2x + 3 (x

^{2}– log x (2x))= 2x + 3 (x – 2x . log x)

= 2x + 3x – 6x . log x

= x(5 – 6 log x)

### Question 6. e^{x }sin 5x

**Solution:**

Here, y = e

^{x}sin 5x

First derivative,Using product rule

= e

^{x}+ sin 5x= e

^{x}(5 cos(5x))+ sin 5x (e^{x})= e

^{x}(5 cos(5x)+ sin 5x)

Second derivative,=

Using product rule,

= e

^{x}+ (5 cos(5x)+ sin 5x)= e

^{x}(5 (5(- sin 5x))) + 5(cos 5x) + (5 cos(5x)+ sin 5x) (e^{x})= e

^{x}(- 25 sin 5x + 5cos 5x) + (5 cos(5x)+ sin 5x) (e^{x})= e

^{x}(- 25 sin 5x + 5cos 5x + 5 cos(5x)+ sin 5x)

= e^{x}(10 cos 5x – 24 sin 5x)

### Question 7. e^{6x} cos 3x

**Solution:**

Here, y = e

^{6x}cos 3x

First derivative,Using product rule

= e

^{6x}+ cos 3x= e

^{6x}(- 3 sin(3x))+ cos 3x (6e^{6x})= e

^{6x}(6 cos(3x) – 3 sin (3x))

Second derivative,=

Using product rule,

= e

^{6x}() + (6 cos(3x) – 3 sin (3x))= e

^{6x}(6 (3 (- sin(3x)) – 3 (3 cos 3x)) + (6 cos(3x) – 3 sin (3x)) (6e^{6x})= e

^{6x}(- 18sin(3x) – 9 cos 3x) + (36 cos(3x) – 18 sin (3x)) (e^{6x})= e

^{6x}(27 cos(3x) – 36 sin (3x))

= 9e^{6x}(3 cos(3x) – 4 sin (3x))

### Question 8. tan^{–1} x

**Solution:**

Here, y = tan

^{–1}x

First derivative,=

Second derivative,=

Using division rule,

=

=

=

### Question 9. log (log x)

**Solution:**

Here, y = log (log x)

First derivative,=

=

=

Second derivative,=

Using division rule,

=

Using product rule,

=

= –

= –

= –

### Question 10. sin (log x)

**Solution:**

Here, y = sin (log x)

First derivative,= cos (log x)

= cos (log x) .

=

Second derivative,=

Using division rule,

=

=

=

=

### Question 11. If y = 5 cos x – 3 sin x, prove that + y = 0

**Solution:**

Here, y = 5 cos x – 3 sin x

First derivative,= 5 (- sin x) – 3 (cos x)

= – 5 sin(x) – 3 cos(x)

Second derivative,=

=

= -5 (cos(x)) – 3 (- sin(x))

= -(5 cos(x) – 3 sin(x))

= -y

According to the given condition,

+ y = -y + y

+ y = 0Hence Proved!!

### Question 12. If y = cos^{-1} x, Find in terms of y alone.

**Solution:**

Here, y = cos-1 x

x = cos y

First derivative,= – sin y

= – cosec (y)

Second derivative,=

= – (-cosec(y) cot (y))

= – (-cosec(y) cot (y)) (-cosec(y))

= -cosec

^{2}(y) cot (y)Hence we get

= -cosec^{2}(y) cot (y)

### Question 13. If y = 3 cos (log x) + 4 sin (log x), show that x^{2} y_{2}+ xy_{1}+ y = 0

**Solution:**

Here, y = 3 cos (log x) + 4 sin (log x)

First derivative,y

_{1}== 3 (-sin (log x)) + 4 (cos (log(x)))

= (4 cos (log(x)-3 sin (log x))

Second derivative,y

_{2}==

Using product rule.

=

= (4(-sin(log(x))) – 3 (cos(log(x)))) + (4 cos (log(x)-3 sin (log x)) ()

= (-4sin(log(x)) – 3 cos(log(x))) – (4 cos (log(x) + 3 sin (log x)) ()

= \frac{-1}{x^2} [-7 cos(log(x) – sin (log x)]

According to the given conditions,

xy

_{1}= x( (4 cos (log(x)-3 sin (log x)))xy

_{1}= -3 sin (log x)+ 4 cos (log(x))x

^{2}y_{2}= x^{2}x

^{2}y_{2}=[-7 cos(log(x) – sin (log x)]Now, rearranging

xy

_{1}+ x^{2}y_{2}+ y = -3 sin (log x)+ 4 cos (log(x)) + cos(log(x)) -7 cos(log(x) – sin (log x) + 4 sin (log x)Hence we get

xy_{1}+ x^{2}y_{2}+ y = 0

### Question 14. If y = Ae^{mx} + Be^{nx}, show that – (m+n) + mny = 0.

**Solution:**

Here, y = Ae

^{mx}+ Be^{nx}

First derivative,= mAe

^{mx}+ nBe^{nx}

Second derivative,=

= m

^{2}Ae^{mx}+ n^{2}Be^{nx}According to the given conditions,

– (m+n) + mny, we get

LHS =m^{2}Ae^{mx}+ n^{2}Be^{nx}– (m+n)(mAe^{mx}+ nBe^{nx}) + mny= m

^{2}Ae^{mx}+ n^{2}Be^{nx}– (m^{2}Ae^{mx}+ mnAe^{mx}+ mnBe^{nx}+ n^{2}Be^{nx}) + mny= -(mnAe

^{mx}+ mnBe^{nx}) + mny= -mny + mny

= 0

Hence we get

+ mny = 0

### Question 15. If y = 500e^{7x}+ 600e^{– 7x}, show that = 49y.

**Solution:**

Here, y = 500e

^{7x}+ 600e^{– 7x}

First derivative,= 500e

^{7x}. (7)+ 600e^{– 7x}(-7)= 7(500e

^{7x}– 600e^{– 7x})

Second derivative,=

= 7[500e

^{7x}. (7) – 600e^{– 7x}. (-7)]= 49[500e

^{7x}+ 600e^{– 7x}]

= 49yHence Proved!!

### Question 16. If e^{y} (x + 1) = 1, show that =

**Solution:**

e

^{y}(x + 1) = 1e

^{-y}= (x+1)

First derivative,-e

^{-y}= 1=

Second derivative,=

Using division rule,

=

=

=

=

Hence we can conclude that,

### Question 17. If y = (tan^{–1} x)^{2}, show that (x^{2}+ 1)^{2} y_{2}+ 2x (x^{2}+ 1) y_{1}= 2

**Solution:**

Here, y = (tan–1 x)

^{2}= 2 . tan–1 x

(x

^{2}+ 1) = 2 tan–1 xDerivation further,

(x

^{2}+ 1) + (x2 + 1) =(x

^{2}+ 1) + (2x) = 2Multiplying (x

^{2}+ 1),(x

^{2}+ 1)^{2}+ (2x)(x^{2}+ 1) = 2Hence Proved,

(x^{2}+ 1)^{2}y_{2}+ 2x (x^{2}+ 1) y_{1}= 2