GSEB Class 11 Maths Solutions Chapter 13 Limits and Derivatives Exercise 13.2

Question 1. Find the derivative of \( x^2 – 2 \) at \( x = 10 \).
Answer: The derivative of \( f(x) \) at \( x = a \) is given by the formula:
\( f'(a) = \lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \)
Here, the function is \( f(x) = x^2 – 2 \). We need to find the derivative at \( a = 10 \).
First, we find \( f(10 + h) = (10 + h)^2 – 2 \).
Next, we find \( f(10) = 10^2 – 2 \).
Now, we subtract \( f(10) \) from \( f(10+h) \):
\( f(10 + h) – f(10) = [(10 + h)^2 – 2] – (10^2 – 2) \)
\( = (10 + h)^2 – 10^2 \)
\( = (100 + 20h + h^2) - 100 \)
\( = 20h + h^2 \)
\( = (20 + h)h \)
Now, substitute this back into the limit definition:
\( f'(10) = \lim _{h \rightarrow 0} \frac{f(10+h)-f(10)}{h} \)
\( = \lim _{h \rightarrow 0} \frac{h(20+h)}{h} \)
Since \( h \rightarrow 0 \), we can cancel \( h \) from the numerator and denominator:
\( = \lim _{h \rightarrow 0} (20 + h) \)
Now, apply the limit as \( h \) approaches 0:
\( = 20 + 0 \)
\( = 20 \)
In simple words: To find the derivative using the first principle, we put \( x+h \) into the function, then subtract the original function value, divide by \( h \), and finally let \( h \) become zero. After doing the math, we found the derivative at \( x=10 \) is 20.

Exam Tip: Remember the definition of the derivative from first principles, \( f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \). Show all algebraic simplification steps clearly before applying the limit to avoid calculation errors.

Question 2. Find the derivative of \( 99x \) at \( x = 100 \).
Answer: The derivative of \( f(x) \) at \( x = a \) is given by the formula:
\( f'(a) = \lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \)
Here, the function is \( f(x) = 99x \). We need to find the derivative at \( a = 100 \).
First, we find \( f(100 + h) = 99(100 + h) \).
Next, we find \( f(100) = 99 \times 100 \).
Now, we subtract \( f(100) \) from \( f(100+h) \):
\( f(100 + h) – f(100) = 99(100 + h) – 99 \times 100 \)
\( = 99[(100 + h) - 100] \)
\( = 99h \)
Now, substitute this back into the limit definition:
\( f'(100) = \lim _{h \rightarrow 0} \frac{99h}{h} \)
Since \( h \rightarrow 0 \), we can cancel \( h \) from the numerator and denominator:
\( = \lim _{h \rightarrow 0} 99 \)
The limit of a constant is the constant itself:
\( = 99 \)
In simple words: We used the first principle formula to find how the function \( 99x \) changes. By substituting and simplifying, we found that its rate of change at \( x=100 \) is always 99.

Exam Tip: For linear functions like \( f(x) = mx + c \), the derivative is always the slope \( m \). In this case, \( f(x) = 99x \) (where \( c = 0 \)), so the derivative is 99, irrespective of the value of \( x \).

Question 3. Find the derivative of \( x \) at \( x = 1 \).
Answer: The derivative of \( f(x) \) at \( x = a \) is given by the formula:
\( f'(a) = \lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h} \)
Here, the function is \( f(x) = x \). We need to find the derivative at \( a = 1 \).
First, we find \( f(1 + h) = (1 + h) \).
Next, we find \( f(1) = 1 \).
Now, we subtract \( f(1) \) from \( f(1+h) \):
\( f(1 + h) – f(1) = (1 + h) – 1 \)
\( = h \)
Now, substitute this back into the limit definition:
\( f'(1) = \lim _{h \rightarrow 0} \frac{h}{h} \)
Since \( h \rightarrow 0 \), we can cancel \( h \) from the numerator and denominator:
\( = \lim _{h \rightarrow 0} 1 \)
The limit of a constant is the constant itself:
\( = 1 \)
In simple words: Using the definition of a derivative, we plug in \( x+h \) and \( x \) into the function, subtract, divide by \( h \), and then take the limit as \( h \) approaches zero. For \( f(x) = x \), the derivative at \( x=1 \) is 1.

Exam Tip: The derivative of \( f(x) = x \) is always 1, regardless of the point \( x \) where it is evaluated. This is a fundamental result in differentiation.

Question 4. Find the derivative of the following functions from first principles:
(i) \( x^3 – 27 \)
(ii) \( (x – 1)(x – 2) \)
(iii) \( \frac{1}{x^{2}} \)
(iv) \( \frac{x+1}{x-1} \)
Answer:
(i) Let \( f(x) = x^3 – 27 \). The derivative \( f'(x) \) by first principle is given by:
\( f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim _{h \rightarrow 0} \frac{[(x+h)^3 - 27] - (x^3 - 27)}{h} \)
\( = \lim _{h \rightarrow 0} \frac{(x+h)^3 - x^3}{h} \)
Expand \( (x+h)^3 \):
\( = \lim _{h \rightarrow 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h} \)
\( = \lim _{h \rightarrow 0} \frac{3x^2h + 3xh^2 + h^3}{h} \)
Factor out \( h \) from the numerator:
\( = \lim _{h \rightarrow 0} \frac{h(3x^2 + 3xh + h^2)}{h} \)
Cancel \( h \) (since \( h \rightarrow 0 \), \( h \neq 0 \)):
\( = \lim _{h \rightarrow 0} (3x^2 + 3xh + h^2) \)
Apply the limit as \( h \) approaches 0:
\( = 3x^2 + 3x(0) + (0)^2 \)
\( = 3x^2 \)
Thus, the derivative of \( x^3 - 27 \) is \( 3x^2 \).

(ii) Let \( f(x) = (x – 1)(x – 2) \). First, expand the function:
\( f(x) = x^2 – 2x – x + 2 = x^2 – 3x + 2 \).
The derivative \( f'(x) \) by first principle is given by:
\( f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim _{h \rightarrow 0} \frac{[(x+h)^2 - 3(x+h) + 2] - (x^2 - 3x + 2)}{h} \)
Expand and simplify the numerator:
\( = \lim _{h \rightarrow 0} \frac{(x^2 + 2xh + h^2 - 3x - 3h + 2) - (x^2 - 3x + 2)}{h} \)
\( = \lim _{h \rightarrow 0} \frac{x^2 + 2xh + h^2 - 3x - 3h + 2 - x^2 + 3x - 2}{h} \)
\( = \lim _{h \rightarrow 0} \frac{2xh + h^2 - 3h}{h} \)
Factor out \( h \) from the numerator:
\( = \lim _{h \rightarrow 0} \frac{h(2x + h - 3)}{h} \)
Cancel \( h \):
\( = \lim _{h \rightarrow 0} (2x + h - 3) \)
Apply the limit as \( h \) approaches 0:
\( = 2x + 0 - 3 \)
\( = 2x - 3 \)
Thus, the derivative of \( (x-1)(x-2) \) is \( 2x-3 \).

(iii) Let \( f(x) = \frac{1}{x^{2}} \). The derivative \( f'(x) \) by first principle is given by:
\( f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim _{h \rightarrow 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} \)
Find a common denominator for the numerator:
\( = \lim _{h \rightarrow 0} \frac{\frac{x^2 - (x+h)^2}{x^2(x+h)^2}}{h} \)
\( = \lim _{h \rightarrow 0} \frac{x^2 - (x+h)^2}{hx^2(x+h)^2} \)
Expand \( (x+h)^2 \):
\( = \lim _{h \rightarrow 0} \frac{x^2 - (x^2 + 2xh + h^2)}{hx^2(x+h)^2} \)
\( = \lim _{h \rightarrow 0} \frac{x^2 - x^2 - 2xh - h^2}{hx^2(x+h)^2} \)
\( = \lim _{h \rightarrow 0} \frac{-2xh - h^2}{hx^2(x+h)^2} \)
Factor out \( h \) from the numerator:
\( = \lim _{h \rightarrow 0} \frac{h(-2x - h)}{hx^2(x+h)^2} \)
Cancel \( h \):
\( = \lim _{h \rightarrow 0} \frac{-2x - h}{x^2(x+h)^2} \)
Apply the limit as \( h \) approaches 0:
\( = \frac{-2x - 0}{x^2(x+0)^2} \)
\( = \frac{-2x}{x^2 \cdot x^2} \)
\( = \frac{-2x}{x^4} \)
\( = -\frac{2}{x^3} \)
Thus, the derivative of \( \frac{1}{x^2} \) is \( -\frac{2}{x^3} \).

(iv) Let \( f(x) = \frac{x+1}{x-1} \). The derivative \( f'(x) \) by first principle is given by:
\( f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim _{h \rightarrow 0} \frac{\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}}{h} \)
Find a common denominator for the numerator:
\( = \lim _{h \rightarrow 0} \frac{\frac{(x-1)(x+h+1) - (x+1)(x+h-1)}{(x+h-1)(x-1)}}{h} \)
\( = \lim _{h \rightarrow 0} \frac{(x-1)(x+h+1) - (x+1)(x+h-1)}{h(x+h-1)(x-1)} \)
Expand the terms in the numerator:
\( (x-1)(x+h+1) = x(x+h+1) - 1(x+h+1) = x^2+xh+x - x-h-1 = x^2+xh-h-1 \)
\( (x+1)(x+h-1) = x(x+h-1) + 1(x+h-1) = x^2+xh-x + x+h-1 = x^2+xh+h-1 \)
Subtract the expanded terms:
\( (x^2+xh-h-1) - (x^2+xh+h-1) \)
\( = x^2+xh-h-1 - x^2-xh-h+1 \)
\( = -2h \)
Substitute this back into the limit expression:
\( = \lim _{h \rightarrow 0} \frac{-2h}{h(x+h-1)(x-1)} \)
Cancel \( h \):
\( = \lim _{h \rightarrow 0} \frac{-2}{(x+h-1)(x-1)} \)
Apply the limit as \( h \) approaches 0:
\( = \frac{-2}{(x+0-1)(x-1)} \)
\( = \frac{-2}{(x-1)(x-1)} \)
\( = \frac{-2}{(x-1)^2} \)
Thus, the derivative of \( \frac{x+1}{x-1} \) is \( \frac{-2}{(x-1)^2} \).
In simple words: For each function, we applied the first principles definition of the derivative. This involved writing the function at \( x+h \), subtracting the original function, dividing by \( h \), and then simplifying the expression before taking the limit as \( h \) goes to zero. This process helps us find the general formula for the rate of change of each function.

Exam Tip: Always expand and simplify the numerator completely before attempting to cancel \( h \) from the denominator. Pay attention to algebraic signs, especially when expanding binomials and distributing negative signs.

Question 5. For the function \( f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + ... + \frac{x^{2}}{2} + x \), prove that \( f'(1) = 100f'(0) \).
Answer: Given the function:
\( f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + ... + \frac{x^{2}}{2} + x \)
First, we find the derivative \( f'(x) \) by applying the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) to each term:
\( f'(x) = \frac{100x^{99}}{100} + \frac{99x^{98}}{99} + ... + \frac{2x}{2} + 1 \)
Simplify each term:
\( f'(x) = x^{99} + x^{98} + ... + x + 1 \)
Now, we evaluate \( f'(x) \) at \( x = 1 \):
\( f'(1) = 1^{99} + 1^{98} + ... + 1 + 1 \)
Since there are 100 terms (from \( x^0 \) to \( x^{99} \)), each being 1:
\( f'(1) = 1 + 1 + ... + 1 \) (100 terms)
\( f'(1) = 100 \)
Next, we evaluate \( f'(x) \) at \( x = 0 \):
\( f'(0) = 0^{99} + 0^{98} + ... + 0 + 1 \)
All terms except the last constant term (1) become 0:
\( f'(0) = 0 + 0 + ... + 0 + 1 \)
\( f'(0) = 1 \)
Now, we need to prove that \( f'(1) = 100f'(0) \).
Substitute the values we found:
\( 100 = 100 \times 1 \)
\( 100 = 100 \)
This statement is true.
Therefore, it is proven that \( f'(1) = 100f'(0) \).
In simple words: We first found the derivative of the given sum of powers using basic differentiation rules. Then, we calculated the value of this derivative at \( x=1 \) and at \( x=0 \). By comparing these two values, we showed that the derivative at 1 is 100 times the derivative at 0, confirming the statement.

Exam Tip: When differentiating a sum of terms, remember to differentiate each term separately. Also, be careful when evaluating at \( x=0 \) as terms with \( x \) as a factor will become zero, leaving only constant terms.

Question 6. Find the derivative of \( x^n + ax^{n-1} + a^2x^{n-2} + ..... + a^{n-1}x + a^n \) for some fixed real number.
Answer: Let \( f(x) = x^n + ax^{n-1} + a^2x^{n-2} + ..... + a^{n-1}x + a^n \).
To find the derivative, we apply the power rule \( \frac{d}{dx}(cx^k) = ckx^{k-1} \) to each term. Remember that \( a \) is a constant, and the derivative of a constant term is 0.
Derivative of \( x^n \) is \( nx^{n-1} \).
Derivative of \( ax^{n-1} \) is \( a(n-1)x^{n-2} \).
Derivative of \( a^2x^{n-2} \) is \( a^2(n-2)x^{n-3} \).
...
Derivative of \( a^{n-1}x \) is \( a^{n-1}(1)x^{1-1} = a^{n-1}x^0 = a^{n-1} \).
Derivative of \( a^n \) (which is a constant) is \( 0 \).
Combining these, the derivative \( f'(x) \) is:
\( f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + ..... + a^{n-1} \).
In simple words: We need to find the derivative of a polynomial where each term has a power of \( x \) multiplied by a constant \( a \). We use the power rule for each term and remember that the derivative of a constant term at the end is zero. This gives us a new polynomial as the derivative.

Exam Tip: Be careful with the coefficients and powers when differentiating. The constant \( a \) should be treated as a numerical multiplier. The derivative of the last term \( a^n \) is 0 because \( a^n \) is a constant.

Question 7. For some constants \( a \) and \( b \), find the derivative of:
(i) \( (x – a)(x – b) \)
(ii) \( (ax^2 + b)^2 \)
(iii) \( \frac{x-a}{x-b} \)
Answer:
(i) Let \( f(x) = (x – a)(x – b) \).
We can either expand the product first or use the product rule. Let's use the product rule \( (uv)' = u'v + uv' \).
Let \( u = (x-a) \) and \( v = (x-b) \).
Then \( u' = \frac{d}{dx}(x-a) = 1 - 0 = 1 \).
And \( v' = \frac{d}{dx}(x-b) = 1 - 0 = 1 \).
So, \( f'(x) = (1)(x-b) + (x-a)(1) \)
\( = x - b + x - a \)
\( = 2x - a - b \)
Alternatively, expand first: \( f(x) = x^2 - bx - ax + ab = x^2 - (a+b)x + ab \).
Then \( f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}((a+b)x) + \frac{d}{dx}(ab) \)
\( = 2x - (a+b)(1) + 0 \)
\( = 2x - a - b \)

(ii) Let \( f(x) = (ax^2 + b)^2 \).
We can either expand the expression first or use the chain rule. Let's expand it first:
\( f(x) = (ax^2)^2 + 2(ax^2)(b) + b^2 \)
\( f(x) = a^2x^4 + 2abx^2 + b^2 \)
Now, differentiate term by term using the power rule:
\( f'(x) = \frac{d}{dx}(a^2x^4) + \frac{d}{dx}(2abx^2) + \frac{d}{dx}(b^2) \)
\( = a^2(4x^3) + 2ab(2x) + 0 \)
\( = 4a^2x^3 + 4abx \)

(iii) Let \( f(x) = \frac{x-a}{x-b} \).
We use the quotient rule \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).
Let \( u = (x-a) \) and \( v = (x-b) \).
Then \( u' = \frac{d}{dx}(x-a) = 1 \).
And \( v' = \frac{d}{dx}(x-b) = 1 \).
So, \( f'(x) = \frac{(1)(x-b) - (x-a)(1)}{(x-b)^2} \)
\( = \frac{x-b - (x-a)}{(x-b)^2} \)
\( = \frac{x-b - x+a}{(x-b)^2} \)
\( = \frac{a-b}{(x-b)^2} \)
In simple words: We used different differentiation rules for each function. For the product of two simple linear terms, we can either expand or use the product rule. For the squared term, expanding it into a polynomial and then differentiating is straightforward. For the fraction, the quotient rule is essential to find the derivative.

Exam Tip: Choose the most efficient differentiation rule. For products, expanding can sometimes be simpler than the product rule if terms are basic. Always be careful with the quotient rule, as errors in algebraic simplification in the numerator are common.

Question 8. Find the derivative of \( \frac{x^{n}-a^{n}}{x-a} \) for some constant \( a \).
Answer: Let \( f(x) = \frac{x^{n}-a^{n}}{x-a} \).
We use the quotient rule \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).
Let \( u = x^n - a^n \)
Then \( u' = \frac{d}{dx}(x^n - a^n) = nx^{n-1} - 0 = nx^{n-1} \) (since \( a^n \) is a constant).
Let \( v = x-a \)
Then \( v' = \frac{d}{dx}(x-a) = 1 - 0 = 1 \).
Now, substitute these into the quotient rule formula:
\( f'(x) = \frac{(nx^{n-1})(x-a) - (x^n-a^n)(1)}{(x-a)^2} \)
Expand the numerator:
\( = \frac{nx^{n-1}x - nx^{n-1}a - x^n + a^n}{(x-a)^2} \)
\( = \frac{nx^n - nax^{n-1} - x^n + a^n}{(x-a)^2} \)
Group the terms with \( x^n \):
\( = \frac{(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2} \)
In simple words: To find the derivative of this fraction, we used the quotient rule. We differentiated the numerator and the denominator separately, then put them into the quotient rule formula. After expanding and grouping similar terms, we got the final derivative.

Exam Tip: The quotient rule is crucial for differentiating rational functions. Be very careful with the algebra in the numerator, especially when distributing terms and combining like terms. Treat \( a \) as a constant, so \( a^n \) and \( a \) differentiate to zero or act as multipliers.

Question 9. Find the derivative of
(i) \( 2x - \frac{3}{4} \)
(ii) \( (5x^3 + 3x – 1)(x – 1) \)
(iii) \( x^{-3}(5 + 3x) \)
(iv) \( x^5(3 – 6x^{-9}) \)
(v) \( x^{-4}(3 – 4x^{-5}) \)
(vi) \( \frac{2}{x+1} - \frac{x^{2}}{3 x-1} \)
Answer:
(i) Let \( f(x) = 2x - \frac{3}{4} \).
Apply the power rule and constant rule:
\( f'(x) = \frac{d}{dx}(2x) - \frac{d}{dx}(\frac{3}{4}) \)
\( = 2(1) - 0 \)
\( = 2 \)

(ii) Let \( f(x) = (5x^3 + 3x – 1)(x – 1) \).
Use the product rule \( (uv)' = u'v + uv' \).
Let \( u = 5x^3 + 3x – 1 \), so \( u' = 15x^2 + 3 \).
Let \( v = x – 1 \), so \( v' = 1 \).
\( f'(x) = (15x^2 + 3)(x – 1) + (5x^3 + 3x – 1)(1) \)
Expand and simplify:
\( = (15x^3 - 15x^2 + 3x - 3) + (5x^3 + 3x - 1) \)
\( = 15x^3 + 5x^3 - 15x^2 + 3x + 3x - 3 - 1 \)
\( = 20x^3 - 15x^2 + 6x - 4 \)

(iii) Let \( f(x) = x^{-3}(5 + 3x) \).
First, expand the function: \( f(x) = 5x^{-3} + 3x^{-3}x = 5x^{-3} + 3x^{-2} \).
Now, differentiate term by term using the power rule:
\( f'(x) = \frac{d}{dx}(5x^{-3}) + \frac{d}{dx}(3x^{-2}) \)
\( = 5(-3)x^{-3-1} + 3(-2)x^{-2-1} \)
\( = -15x^{-4} - 6x^{-3} \)
To write with positive exponents:
\( = -\frac{15}{x^4} - \frac{6}{x^3} \)
Combine with a common denominator:
\( = \frac{-15 - 6x}{x^4} = -\frac{3(5 + 2x)}{x^4} \)

(iv) Let \( f(x) = x^5(3 – 6x^{-9}) \).
First, expand the function: \( f(x) = 3x^5 - 6x^5x^{-9} = 3x^5 - 6x^{5-9} = 3x^5 - 6x^{-4} \).
Now, differentiate term by term:
\( f'(x) = \frac{d}{dx}(3x^5) - \frac{d}{dx}(6x^{-4}) \)
\( = 3(5)x^{5-1} - 6(-4)x^{-4-1} \)
\( = 15x^4 + 24x^{-5} \)
To write with positive exponents:
\( = 15x^4 + \frac{24}{x^5} \)
Combine with a common denominator:
\( = \frac{15x^9 + 24}{x^5} = \frac{3(5x^9 + 8)}{x^5} \)

(v) Let \( f(x) = x^{-4}(3 – 4x^{-5}) \).
First, expand the function: \( f(x) = 3x^{-4} - 4x^{-4}x^{-5} = 3x^{-4} - 4x^{-4-5} = 3x^{-4} - 4x^{-9} \).
Now, differentiate term by term:
\( f'(x) = \frac{d}{dx}(3x^{-4}) - \frac{d}{dx}(4x^{-9}) \)
\( = 3(-4)x^{-4-1} - 4(-9)x^{-9-1} \)
\( = -12x^{-5} + 36x^{-10} \)
To write with positive exponents:
\( = -\frac{12}{x^5} + \frac{36}{x^{10}} \)
Combine with a common denominator:
\( = \frac{-12x^5 + 36}{x^{10}} = \frac{12(3 - x^5)}{x^{10}} \)

(vi) Let \( f(x) = \frac{2}{x+1} - \frac{x^{2}}{3x-1} \).
We need to differentiate each term separately using the quotient rule.
For the first term, let \( f_1(x) = \frac{2}{x+1} \). Let \( u_1=2 \), \( v_1=x+1 \). So \( u_1'=0 \), \( v_1'=1 \).
\( f_1'(x) = \frac{(0)(x+1) - (2)(1)}{(x+1)^2} = \frac{-2}{(x+1)^2} \)
For the second term, let \( f_2(x) = \frac{x^{2}}{3x-1} \). Let \( u_2=x^2 \), \( v_2=3x-1 \). So \( u_2'=2x \), \( v_2'=3 \).
\( f_2'(x) = \frac{(2x)(3x-1) - (x^2)(3)}{(3x-1)^2} \)
\( = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} \)
\( = \frac{3x^2 - 2x}{(3x-1)^2} \)
Now, combine the derivatives:
\( f'(x) = f_1'(x) - f_2'(x) \)
\( = \frac{-2}{(x+1)^2} - \frac{3x^2 - 2x}{(3x-1)^2} \)
In simple words: We found the derivative for each function. For simple sums, we differentiate term by term. For products like in (ii), we use the product rule. For functions that can be expanded into polynomials like in (iii), (iv), and (v), it's often easier to expand first and then differentiate each term. For fractions like in (vi), we apply the quotient rule to each fractional part.

Exam Tip: Simplify the function algebraically before differentiating whenever possible (e.g., expand products or combine terms with exponents). This often makes differentiation easier and less prone to errors than directly applying complex rules like the product or quotient rule.

Question 10. Find the derivative of \( \cos x \) from the first principle.
Answer: Let \( f(x) = \cos x \).
The derivative \( f'(x) \) from first principles is given by:
\( f'(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \)
\( = \lim _{h \rightarrow 0} \frac{\cos(x+h) - \cos x}{h} \)
Use the trigonometric identity: \( \cos A - \cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2}) \).
Here, \( A = x+h \) and \( B = x \).
\( \frac{A+B}{2} = \frac{x+h+x}{2} = \frac{2x+h}{2} = x + \frac{h}{2} \)
\( \frac{A-B}{2} = \frac{x+h-x}{2} = \frac{h}{2} \)
Substitute this into the expression:
\( = \lim _{h \rightarrow 0} \frac{-2\sin(x + \frac{h}{2})\sin(\frac{h}{2})}{h} \)
Rearrange the terms to use the special limit \( \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \):
\( = \lim _{h \rightarrow 0} \left[ -2\sin(x + \frac{h}{2}) \cdot \frac{\sin(\frac{h}{2})}{h} \right] \)
To match the form \( \frac{\sin \theta}{\theta} \), we need \( \frac{h}{2} \) in the denominator. Multiply and divide by 2:
\( = \lim _{h \rightarrow 0} \left[ -\sin(x + \frac{h}{2}) \cdot \frac{\sin(\frac{h}{2})}{(\frac{h}{2})} \right] \)
Now apply the limit:
As \( h \rightarrow 0 \), \( \frac{h}{2} \rightarrow 0 \).
\( \lim _{h \rightarrow 0} \sin(x + \frac{h}{2}) = \sin(x + 0) = \sin x \)
\( \lim _{h \rightarrow 0} \frac{\sin(\frac{h}{2})}{(\frac{h}{2})} = 1 \)
So, \( f'(x) = -\sin x \cdot 1 \)
\( f'(x) = -\sin x \)
In simple words: To find the derivative of \( \cos x \) using the first principle, we used the limit definition. We applied a trigonometric identity to simplify the cosine difference, then rearranged the terms to use the known limit for \( \frac{\sin \theta}{\theta} \). Finally, by applying the limit, we found that the derivative is \( -\sin x \).

Exam Tip: Mastering trigonometric identities, especially sum-to-product formulas, is crucial for derivatives from first principles involving trigonometric functions. Also, remember the fundamental limit \( \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \).

Question 11. Find the derivative of the following functions:
(i) \( \sin x \cos x \)
(ii) \( \sec x \)
(iii) \( 5\sec x + 4\cos x \)
(iv) \( \operatorname{cosec} x \)
(v) \( 3\cot x + 5\operatorname{cosec} x \)
(vi) \( 5\sin x - 6\cos x + 7 \)
(vii) \( 2\tan x - 7\sec x \)
Answer:
(i) Let \( f(x) = \sin x \cos x \).
Use the product rule: \( (uv)' = u'v + uv' \).
Let \( u = \sin x \), so \( u' = \cos x \).
Let \( v = \cos x \), so \( v' = -\sin x \).
\( f'(x) = (\cos x)(\cos x) + (\sin x)(-\sin x) \)
\( = \cos^2 x - \sin^2 x \)
Using the double angle identity \( \cos(2x) = \cos^2 x - \sin^2 x \):
\( f'(x) = \cos 2x \)

(ii) Let \( f(x) = \sec x \).
We know that \( \sec x = \frac{1}{\cos x} \). Use the quotient rule \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).
Let \( u = 1 \), so \( u' = 0 \).
Let \( v = \cos x \), so \( v' = -\sin x \).
\( f'(x) = \frac{(0)(\cos x) - (1)(-\sin x)}{\cos^2 x} \)
\( = \frac{\sin x}{\cos^2 x} \)
\( = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \)
\( = \tan x \sec x \)

(iii) Let \( f(x) = 5\sec x + 4\cos x \).
Differentiate each term separately. We know \( \frac{d}{dx}(\sec x) = \sec x \tan x \) and \( \frac{d}{dx}(\cos x) = -\sin x \).
\( f'(x) = 5\frac{d}{dx}(\sec x) + 4\frac{d}{dx}(\cos x) \)
\( = 5(\sec x \tan x) + 4(-\sin x) \)
\( = 5\sec x \tan x - 4\sin x \)

(iv) Let \( f(x) = \operatorname{cosec} x \).
We know that \( \operatorname{cosec} x = \frac{1}{\sin x} \). Use the quotient rule.
Let \( u = 1 \), so \( u' = 0 \).
Let \( v = \sin x \), so \( v' = \cos x \).
\( f'(x) = \frac{(0)(\sin x) - (1)(\cos x)}{\sin^2 x} \)
\( = \frac{-\cos x}{\sin^2 x} \)
\( = -\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} \)
\( = -\cot x \operatorname{cosec} x \)

(v) Let \( f(x) = 3\cot x + 5\operatorname{cosec} x \).
Differentiate each term separately. We know \( \frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x \) and \( \frac{d}{dx}(\operatorname{cosec} x) = -\operatorname{cosec} x \cot x \).
\( f'(x) = 3\frac{d}{dx}(\cot x) + 5\frac{d}{dx}(\operatorname{cosec} x) \)
\( = 3(-\operatorname{cosec}^2 x) + 5(-\operatorname{cosec} x \cot x) \)
\( = -3\operatorname{cosec}^2 x - 5\operatorname{cosec} x \cot x \)
Factor out \( -\operatorname{cosec} x \):
\( = -\operatorname{cosec} x (3\operatorname{cosec} x + 5\cot x) \)

(vi) Let \( f(x) = 5\sin x - 6\cos x + 7 \).
Differentiate each term separately. We know \( \frac{d}{dx}(\sin x) = \cos x \) and \( \frac{d}{dx}(\cos x) = -\sin x \). The derivative of a constant is 0.
\( f'(x) = 5\frac{d}{dx}(\sin x) - 6\frac{d}{dx}(\cos x) + \frac{d}{dx}(7) \)
\( = 5(\cos x) - 6(-\sin x) + 0 \)
\( = 5\cos x + 6\sin x \)

(vii) Let \( f(x) = 2\tan x - 7\sec x \).
Differentiate each term separately. We know \( \frac{d}{dx}(\tan x) = \sec^2 x \) and \( \frac{d}{dx}(\sec x) = \sec x \tan x \).
\( f'(x) = 2\frac{d}{dx}(\tan x) - 7\frac{d}{dx}(\sec x) \)
\( = 2(\sec^2 x) - 7(\sec x \tan x) \)
Factor out \( \sec x \):
\( = \sec x (2\sec x - 7\tan x) \)
In simple words: For all these problems, we used standard differentiation rules for trigonometric functions, products, sums, and quotients. For sums and differences, we differentiated each part separately. For products, we applied the product rule. For terms like \( \sec x \) and \( \operatorname{cosec} x \), we can use the quotient rule or recall their direct derivative formulas.

Exam Tip: Memorize the derivatives of all six basic trigonometric functions. This will save significant time and prevent errors compared to deriving them each time using the quotient rule. Be careful with plus and minus signs, especially for co-functions (cosine, cotangent, cosecant) whose derivatives are negative.