GSEB Solutions Class 12 Statistics Part 2 Chapter 5 Differentiation Ex 5.1

Gujarat Board Statistics Class 12 GSEB Solutions Part 2 Chapter 5 Differentiation Ex 5.1 Textbook Exercise Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Statistics Part 2 Chapter 5 Differentiation Ex 5.1

Obtain the derivatives of the following functions with the help of definition:

Question 1.
f(x) = 2x + 3
Solution:
Here, f(x) = 2x + 3
∴ f(x + h) = 2 (x + h) + 3 = 2x + 2h + 3

Hence, f(x) = 2x + 3 then f’(x) = 2.

f(x) = x²
Solution:
Here, f(x) = x²
∴ f(x + h) = (x + h)² = x² + 2xh + h²

Hence, f(x) = x² then f'(x) = 2x.

Question 3.
f(x) = x⁷
Solution:
Here, f(x) = x⁷
∴ f(x + h) = (x + h)⁷

Take, x + h = t, when h → 0, t → x and h = t – x

Hence, f(x) = x⁷ then f'(x) = 7x⁶

Question 4.
f(x) = $\frac{1}{x+1}$, x ≠ -1
Solution:
Here, f(x) = $\frac{1}{x+1}$
∴ f(x + h) = f(x)

Hence, f(x) = $\frac{1}{x+1}$ then f'(x) = $-\frac{1}{(x+1)^{2}}$.

Question 5.
f(x) = $\sqrt[3]{x}$
solution:
Here, f(x) = $\sqrt[3]{x}=x^{\frac{1}{3}}$

∴ f(x + h) = $(x+h)^{\frac{1}{3}}$

Take, x + h = t, when h → 0, t → x and h = t – x

Hence, f(x) = $\sqrt[3]{x}$ then f'(x) = $\frac{1}{3 \cdot x^{\frac{2}{3}}}$.

Question 6.
f(x) = 24, x ≠ $\frac{4}{3}$
Solution:
Here, f(x) = $\frac{2}{3 x-4}$

∴ f(x + h) = $\frac{2}{3(x+h)-4}$

Hence, f(x) = $\frac{2}{3(x+h)-4}$ then f’(x) = $\frac{-6}{(3 x-4)^{2}}$.

Question 7.
f(x) = 10
Solution:
Here, f(x) = 10
∴ f(x + h) = 10

Hence, f(x) = 10 then f’(x) = 0