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

Exercise 5.7 focuses on the application of derivatives in approximation and errors. This exercise explores how derivatives can be used to estimate function values and calculate errors in measurements or approximations. It introduces students to concepts like absolute and relative errors, percentage error, and how these relate to the derivative of a function.

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

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,

= 20x^20-1

= 20x¹⁹

Second derivative,

= 

= 20(19x^19-1)

= 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. 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 = 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 = 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,

= 

= m²Ae^mx + n²Be^nx

According to the given conditions,

 - (m+n) + mny, we get

LHS = m²Ae^mx + n²Be^nx - (m+n)(mAe^mx + nBe^nx) + mny

= m²Ae^mx + n²Be^nx - (m²Ae^mx + mnAe^mx + mnBe^nx + n²Be^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}]

= 49y

Hence Proved!!

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

Solution:

e^y (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 = (tan^–1 x)², show that (x²+ 1)² y₂+ 2x (x²+ 1) y₁= 2

Solution:

Here, y = (tan–1 x)²

= 2 . tan–1 x 

(x² + 1)  = 2 tan–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

Summary

Exercise 5.7 applies the concept of derivatives to problems involving approximations and error calculations. It covers topics such as using differentials to approximate function values, estimating errors in measurements and calculations, and understanding the relationship between small changes in variables and the resulting changes in function values. This exercise helps students understand how calculus can be applied to real-world situations where precise measurements are not always possible.