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# Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.7

### Question 1. x2+ 3x + 2

Solution:

Here, y = x2+ 3x + 2

First derivative,

= 2x+ 3

Second derivative,

= 2

### Question 2. x20

Solution:

Here, y = x20

First derivative,

= 20x20-1

= 20x19

Second derivative,

= 20(19x19-1)

= 380x18

### 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. x3 log x

Solution:

Here, y = x3 . log x

First derivative,

Using product rule

= x3  + log x

= x3 () + log x (3x2)

= x2 + 3x2 log x

Second derivative,

+

Using product rule,

= 2x + 3 (x2  – log x)

= 2x + 3 (x2 – log x (2x))

= 2x + 3 (x – 2x . log x)

= 2x + 3x – 6x . log x

= x(5 – 6 log x)

### Question 6. ex sin 5x

Solution:

Here, y = ex sin 5x

First derivative,

Using product rule

= ex + sin 5x

= ex (5 cos(5x))+ sin 5x (ex)

= ex (5 cos(5x)+ sin 5x)

Second derivative,

=

Using product rule,

= ex  + (5 cos(5x)+ sin 5x)

= ex (5 (5(- sin 5x))) + 5(cos 5x) + (5 cos(5x)+ sin 5x) (ex)

= ex (- 25 sin 5x + 5cos 5x) + (5 cos(5x)+ sin 5x) (ex)

= ex (- 25 sin 5x + 5cos 5x + 5 cos(5x)+ sin 5x)

= ex (10 cos 5x – 24 sin 5x)

### Question 7. e6x cos 3x

Solution:

Here, y = e6x cos 3x

First derivative,

Using product rule

= e6x  + cos 3x

= e6x (- 3 sin(3x))+ cos 3x (6e6x)

= e6x (6 cos(3x) – 3 sin (3x))

Second derivative,

Using product rule,

= e6x () + (6 cos(3x) – 3 sin (3x))

= e6x (6 (3 (- sin(3x)) – 3 (3 cos 3x)) + (6 cos(3x) – 3 sin (3x)) (6e6x)

= e6x (- 18sin(3x) – 9 cos 3x) + (36 cos(3x) – 18 sin (3x)) (e6x)

= e6x (27 cos(3x) – 36 sin (3x))

= 9e6x (3 cos(3x) – 4 sin (3x))

### Question 8. tanâ€“1 x

Solution:

Here, y = tanâ€“1

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 â€“ 3 sin x, prove that  + y = 0

Solution:

Here, y = 5 cos x â€“ 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 = 0

Hence 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))

= -cosec2(y) cot (y)

Hence we get

= -cosec2(y) cot (y)

### Question 13. If y = 3 cos (log x) + 4 sin (log x), show that x2 y2+ xy1+ y = 0

Solution:

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

First derivative,

y1

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

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

Second derivative,

y2

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,

xy1 = x( (4 cos (log(x)-3 sin (log x)))

xy1 = -3 sin (log x)+ 4 cos (log(x))

x2 y2 = x2

x2 y2 =[-7 cos(log(x) – sin (log x)]

Now, rearranging

xy1 +  x2 y2 + 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

xy1 +  x2 y2 + y = 0

### Question 14. If y = Aemx + Benx, show that  – (m+n) + mny = 0.

Solution:

Here, y = Aemx + Benx

First derivative,

= mAemx + nBenx

Second derivative,

= m2Aemx + n2Benx

According to the given conditions,

– (m+n) + mny, we get

LHS = m2Aemx + n2Benx – (m+n)(mAemx + nBenx) + mny

= m2Aemx + n2Benx – (m2Aemx + mnAemx + mnBenx + n2Benx) + mny

= -(mnAemx + mnBenx) + mny

= -mny + mny

= 0

Hence we get

+ mny = 0

### Question 15. If y = 500e7x+ 600eâ€“ 7x, show that  = 49y.

Solution:

Here, y = 500e7x+ 600eâ€“ 7x

First derivative,

= 500e7x . (7)+ 600eâ€“ 7x (-7)

= 7(500e7x – 600eâ€“ 7x)

Second derivative,

= 7[500e7x . (7) – 600eâ€“ 7x . (-7)]

= 49[500e7x + 600eâ€“ 7x]

= 49y

Hence Proved!!

### Question 16. If ey (x + 1) = 1, show that  =

Solution:

ey (x + 1) = 1

e-y = (x+1)

First derivative,

-e-y  = 1

Second derivative,

Using division rule,

Hence we can conclude that,

### Question 17. If y = (tanâ€“1 x)2, show that (x2+ 1)2 y2+ 2x (x2+ 1) y1= 2

Solution:

Here, y = (tanâ€“1 x)2

= 2 . tanâ€“1 x

(x2 + 1)  = 2 tanâ€“1 x

Derivation further,

(x2 + 1) +  (x2 + 1) =

(x2 + 1) +  (2x) = 2

Multiplying (x2 + 1),

(x2 + 1)2 +  (2x)(x2 + 1) = 2

Hence Proved,

(x2+ 1)2 y2+ 2x (x2+ 1) y1= 2

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