GSEB Class 12 Statistics Solutions Chapter 5 Differentiation Exercise 5.1

Gseb 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
Answer: To find the derivative of \(f(x) = 2x + 3\) using the definition, we first identify \(f(x) = 2x + 3\). Then, we find \(f(x + h)\) by replacing \(x\) with \(x + h\), which gives \(f(x + h) = 2(x + h) + 3 = 2x + 2h + 3\). Next, we use the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substitute the expressions for \(f(x+h)\) and \(f(x)\): \[ = \lim_{h \to 0} \frac{(2x + 2h + 3) - (2x + 3)}{h} \] Simplify the numerator: \[ = \lim_{h \to 0} \frac{2x + 2h + 3 - 2x - 3}{h} \] \[ = \lim_{h \to 0} \frac{2h}{h} \] Cancel out \(h\) (since \(h \ne 0\) as \(h \to 0\)): \[ = \lim_{h \to 0} 2 \] The limit of a constant is the constant itself: \[ = 2 \] Thus, if \(f(x) = 2x + 3\), then its derivative \(f'(x) = 2\).
In simple words: We found the slope of the tangent line to the function \(f(x) = 2x + 3\) by using the limit definition of the derivative. After simplifying, we see that the rate of change is a constant 2.

🎯 Exam Tip: Remember to clearly show each step of the limit calculation when using the definition of the derivative, especially simplifying the numerator before taking the limit.

Question 2. f(x) = x\(^2\)
Answer: We want to find the derivative of \(f(x) = x^2\) using the definition. First, we identify \(f(x) = x^2\). Then, we find \(f(x + h)\) by substituting \(x + h\) for \(x\), which gives \(f(x + h) = (x + h)^2 = x^2 + 2xh + h^2\). Now, we apply the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substitute the expressions for \(f(x+h)\) and \(f(x)\): \[ = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2) - x^2}{h} \] Simplify the numerator: \[ = \lim_{h \to 0} \frac{2xh + h^2}{h} \] Factor out \(h\) from the numerator: \[ = \lim_{h \to 0} \frac{h(2x + h)}{h} \] Cancel out \(h\): \[ = \lim_{h \to 0} (2x + h) \] Now, substitute \(h = 0\) into the expression: \[ = 2x + 0 \] \[ = 2x \] Therefore, if \(f(x) = x^2\), its derivative \(f'(x) = 2x\).
In simple words: We found how \(f(x) = x^2\) changes by using limits. After simplifying, the derivative tells us the function's rate of change is \(2x\).

🎯 Exam Tip: When dealing with algebraic expressions like \((x+h)^2\), remember to expand them correctly to avoid errors in simplification.

Question 3. f(x) = x\(^7\)
Answer: To find the derivative of \(f(x) = x^7\), we start by setting \(f(x) = x^7\). Then, \(f(x + h) = (x + h)^7\). Using the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substitute the functions: \[ = \lim_{h \to 0} \frac{(x+h)^7 - x^7}{h} \] To evaluate this limit, we can use a substitution. Let \(x + h = t\). As \(h \to 0\), it implies that \(t \to x\). Also, \(h = t - x\). So the limit becomes: \[ f'(x) = \lim_{t \to x} \frac{t^7 - x^7}{t - x} \] This is a standard form for a limit: \( \lim_{y \to a} \frac{y^n - a^n}{y - a} = n \cdot a^{n-1} \). Applying this rule with \(n = 7\) and \(a = x\): \[ = 7 \cdot x^{7-1} \] \[ = 7x^6 \] Thus, for \(f(x) = x^7\), the derivative \(f'(x) = 7x^6\).
In simple words: We found the derivative of \(x^7\) using the definition. We changed the variables to use a known limit formula, which quickly gave us \(7x^6\) as the answer.

🎯 Exam Tip: Recognizing standard limit forms, like the one used here for \(a^n - x^n\), can save significant time in calculations. Ensure you correctly identify \(n\) and \(a\).

Question 4. f(x) = \(\frac{1}{x+1}\), x \(\neq\) -1
Answer: We need to find the derivative of \(f(x) = \frac{1}{x+1}\). First, \(f(x) = \frac{1}{x+1}\). Then, \(f(x + h) = \frac{1}{(x+h)+1}\). Apply the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substitute the function expressions: \[ = \lim_{h \to 0} \frac{\frac{1}{x+h+1} - \frac{1}{x+1}}{h} \] Combine the fractions in the numerator: \[ = \lim_{h \to 0} \frac{\frac{(x+1) - (x+h+1)}{(x+h+1)(x+1)}}{h} \] Simplify the numerator: \[ = \lim_{h \to 0} \frac{\frac{x+1-x-h-1}{(x+h+1)(x+1)}}{h} \] \[ = \lim_{h \to 0} \frac{\frac{-h}{(x+h+1)(x+1)}}{h} \] Multiply by \(1/h\): \[ = \lim_{h \to 0} \frac{-h}{h(x+h+1)(x+1)} \] Cancel out \(h\): \[ = \lim_{h \to 0} \frac{-1}{(x+h+1)(x+1)} \] Now, substitute \(h = 0\): \[ = \frac{-1}{(x+0+1)(x+1)} \] \[ = \frac{-1}{(x+1)(x+1)} \] \[ = -\frac{1}{(x+1)^2} \] Thus, for \(f(x) = \frac{1}{x+1}\), the derivative \(f'(x) = -\frac{1}{(x+1)^2}\).
In simple words: We found the derivative of \(f(x) = \frac{1}{x+1}\) using the limit method. After combining fractions and simplifying, the derivative came out to be \(-\frac{1}{(x+1)^2}\).

🎯 Exam Tip: When dealing with rational functions, remember to find a common denominator in the numerator before simplifying. Careful algebraic manipulation is key to avoid mistakes.

Question 5. f(x) = \(\sqrt[3]{x}\)
Answer: To find the derivative of \(f(x) = \sqrt[3]{x}\), we first rewrite the function as \(f(x) = x^{\frac{1}{3}}\). Then, \(f(x + h) = (x+h)^{\frac{1}{3}}\). Using the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substitute the functions: \[ = \lim_{h \to 0} \frac{(x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}}{h} \] Similar to Question 3, we can use a substitution. Let \(x + h = t\). As \(h \to 0\), it means \(t \to x\), and \(h = t - x\). The limit becomes: \[ f'(x) = \lim_{t \to x} \frac{t^{\frac{1}{3}} - x^{\frac{1}{3}}}{t - x} \] This matches the standard limit form: \( \lim_{y \to a} \frac{y^n - a^n}{y - a} = n \cdot a^{n-1} \). Here, \(n = \frac{1}{3}\) and \(a = x\). Apply the rule: \[ = \frac{1}{3} \cdot x^{\frac{1}{3}-1} \] Calculate the exponent: \[ = \frac{1}{3} \cdot x^{\frac{1-3}{3}} \] \[ = \frac{1}{3} \cdot x^{-\frac{2}{3}} \] Rewrite with a positive exponent: \[ = \frac{1}{3x^{\frac{2}{3}}} \] Therefore, for \(f(x) = \sqrt[3]{x}\), the derivative \(f'(x) = \frac{1}{3x^{\frac{2}{3}}}\).
In simple words: We found the derivative of the cube root of \(x\) by converting it to a power of \(x\) and then using a special limit rule. This simplified the process to get the derivative as \( \frac{1}{3x^{\frac{2}{3}}} \).

🎯 Exam Tip: Remember to express root functions as fractional exponents before applying the limit definition, as this often simplifies the problem to a standard limit form.

Question 6. f(x) = \(\frac{2}{3x-4}\)
Answer: We need to find the derivative of \(f(x) = \frac{2}{3x-4}\). First, \(f(x) = \frac{2}{3x-4}\). Then, \(f(x + h) = \frac{2}{3(x+h)-4}\). Apply the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substitute the function expressions: \[ = \lim_{h \to 0} \frac{\frac{2}{3(x+h)-4} - \frac{2}{3x-4}}{h} \] Combine the fractions in the numerator: \[ = \lim_{h \to 0} \frac{\frac{2(3x-4) - 2(3(x+h)-4)}{(3(x+h)-4)(3x-4)}}{h} \] Simplify the numerator: \[ = \lim_{h \to 0} \frac{\frac{6x-8 - 2(3x+3h-4)}{(3x+3h-4)(3x-4)}}{h} \] \[ = \lim_{h \to 0} \frac{\frac{6x-8 - 6x-6h+8}{(3x+3h-4)(3x-4)}}{h} \] \[ = \lim_{h \to 0} \frac{\frac{-6h}{(3x+3h-4)(3x-4)}}{h} \] Multiply by \(1/h\): \[ = \lim_{h \to 0} \frac{-6h}{h(3x+3h-4)(3x-4)} \] Cancel out \(h\): \[ = \lim_{h \to 0} \frac{-6}{(3x+3h-4)(3x-4)} \] Now, substitute \(h = 0\): \[ = \frac{-6}{(3x+3(0)-4)(3x-4)} \] \[ = \frac{-6}{(3x-4)(3x-4)} \] \[ = \frac{-6}{(3x-4)^2} \] Thus, for \(f(x) = \frac{2}{3x-4}\), the derivative \(f'(x) = \frac{-6}{(3x-4)^2}\).
In simple words: We used the limit definition to find how the function \(f(x) = \frac{2}{3x-4}\) changes. We combined the fractions and simplified the expression until we could substitute \(h=0\), giving us \( \frac{-6}{(3x-4)^2} \).

🎯 Exam Tip: Be very careful with algebraic signs, especially when distributing negative numbers across parentheses. A small error can lead to an incorrect final derivative.

Question 7. f(x) = 10
Answer: We need to find the derivative of \(f(x) = 10\). First, \(f(x) = 10\). Since 10 is a constant, \(f(x + h)\) will also be 10, because the function does not depend on \(x\). So, \(f(x + h) = 10\). Apply the definition of the derivative: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Substitute the function expressions: \[ = \lim_{h \to 0} \frac{10 - 10}{h} \] Simplify the numerator: \[ = \lim_{h \to 0} \frac{0}{h} \] Since \(h \to 0\) but \(h \neq 0\), the fraction \(\frac{0}{h}\) is always 0. \[ = \lim_{h \to 0} 0 \] The limit of a constant (0) is the constant itself: \[ = 0 \] Therefore, for \(f(x) = 10\), the derivative \(f'(x) = 0\).
In simple words: We found the derivative of the constant function \(f(x) = 10\). Since a constant function never changes its value, its rate of change (derivative) is always zero.

🎯 Exam Tip: Always remember that the derivative of any constant function is zero. This is a fundamental rule in differentiation and a common point for quick marks.