Tamil Nadu Board of Secondary EducationHSC Arts Class 11

# Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1? f(x)=1-x2 - Mathematics

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Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?

f(x) = sqrt(1 - x^2)

#### Solution

f(x) = sqrt(1 - x^2) at x = 1

To find the left limit of f(x) at x = 1

Put x = 1 – h

h > 0

When x → 1

We have h → 0

f"'"(1 - "h") =  lim_("h" -> 0) (f(1 - "h") - f(1))/(1 - "h" - 1)

= lim_("h" -> 0) (sqrt(1 - (1 - "h")^2) - sqrt(1 - 1^2))/(- "h")

= lim_("h" -> 0) (sqrt(1 - (1 - 2"h" + "h"^2)) - 0)/(- "h")

= lim_("h" -> 0) sqrt(1 + 1 + 2"h" - "h"^2)/(- "h")

= lim_("h" -> 0) sqrt("h"^2 (2/"h" - 1))/(- "h")

= lim_("h" -> 0) ("h" sqrt(2/"h" - 1))/(- "h")

= - lim_("h" -> 0) sqrt(2/"h" - 1)

f"'" (1^-) = - sqrt(2/0 - 1) = - sqrt(oo - 1)

f"'"(1^-) = - oo  .........(1)

To find the right limit of f(x) at x = 1

Put x = 1 + h

h > 0

When x → 1

We have h → 0

f"'"(1 + "h") =  lim_("h" -> 0) (f(1 + "h") - f(1))/(1 + "h" - 1)

= lim_("h" -> 0) (sqrt(1 - (1 + "h")^2) - sqrt(1 - 1^2))/"h"

= lim_("h" -> 0) sqrt(1 - (1 + 2"h" + "h"^2))/"h"

= lim_("h" -> 0) sqrt(1 - 1 - 2"h" - "h"^2)/"h"

= lim_("h" -> 0) sqrt(- "h"^2(1 + 2/"h"))/"h"

= lim_("h" -> 0) (sqrt(- 1) * "h")/"h" * sqrt(1 + 2/"h")

= lim_(h' -> 0) "i" * sqrt(1 + 2/"h")

f"'"(1^+) = "i" sqrt(1 + 2/0) = "i" xx oo  .......(2)

From equations (1) and (2) we get

f"'"(1^-)  ≠  f"'"(1^+)

∴ f"'"(x) does not exist at x = 1

Hence f(x) = sqrt(1 - x^2) is n differentiable at x = 1.

Concept: Differentiability and Continuity
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Chapter 10: Differential Calculus - Differentiability and Methods of Differentiation - Exercise 10.1 [Page 147]

#### APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 11th Mathematics Volume 1 and 2 Answers Guide
Chapter 10 Differential Calculus - Differentiability and Methods of Differentiation
Exercise 10.1 | Q 2. (ii) | Page 147
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