OP Malhotra Class 11 Maths Solutions Chapter 19 Differentiation Chapter Test

S Chand Class 11 ICSE Maths Solutions Chapter 19 Differentiation Chapter Test

Question 1. Find from first principles the differential coefficient of \( 2x^2 + 3x \).
Answer: To find the differential coefficient from first principles, we use the definition of the derivative.
Let \( y = f(x) = 2x^2 + 3x \).
Then, \( f(x + \delta x) = 2(x + \delta x)^2 + 3(x + \delta x) \).
Now, we apply the first principle formula for the derivative:
\( \frac{dy}{dx} = f'(x) = \text{Lt}_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \text{Lt}_{\delta x \to 0} \frac{[2(x + \delta x)^2 + 3(x + \delta x)] - [2x^2 + 3x]}{\delta x} \)
\( \implies \frac{dy}{dx} = \text{Lt}_{\delta x \to 0} \frac{[2(x^2 + 2x\delta x + (\delta x)^2) + 3x + 3\delta x] - [2x^2 + 3x]}{\delta x} \)
\( \implies \frac{dy}{dx} = \text{Lt}_{\delta x \to 0} \frac{2x^2 + 4x\delta x + 2(\delta x)^2 + 3x + 3\delta x - 2x^2 - 3x}{\delta x} \)
\( \implies \frac{dy}{dx} = \text{Lt}_{\delta x \to 0} \frac{4x\delta x + 2(\delta x)^2 + 3\delta x}{\delta x} \)
\( \implies \frac{dy}{dx} = \text{Lt}_{\delta x \to 0} (4x + 2\delta x + 3) \)
Now, substitute \( \delta x = 0 \):
\( \implies \frac{dy}{dx} = 4x + 2(0) + 3 \)
\( \implies \frac{dy}{dx} = 4x + 3 \)
In simple words: To find the differential coefficient, we use a formula involving a very tiny change in x, called delta x. We expand the expression, simplify it, and then let delta x become zero to get the final answer. This method is the basic way to find the rate of change of a function.

🎯 Exam Tip: Remember to expand \( (x + \delta x)^2 \) carefully and ensure all terms are correctly cancelled or simplified before taking the limit. A common mistake is simplifying too early or incorrectly.

Question 2. Find from first principles the differential coefficient of \( \sin 2x \).
Answer: We use the first principle definition of differentiation to solve this.
Let \( y = f(x) = \sin 2x \).
Then, \( f(x + \delta x) = \sin 2(x + \delta x) \).
The formula for differentiation from first principles is:
\( \frac{dy}{dx} = f'(x) = \text{Lt}_{\delta x \to 0} \frac{f(x + \delta x) - f(x)}{\delta x} \)
\( \implies \frac{dy}{dx} = \text{Lt}_{\delta x \to 0} \frac{\sin 2(x + \delta x) - \sin 2x}{\delta x} \)
We use the trigonometric identity \( \sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right) \).
Here, \( C = 2(x + \delta x) = 2x + 2\delta x \) and \( D = 2x \).
So, \( \frac{C+D}{2} = \frac{2x + 2\delta x + 2x}{2} = \frac{4x + 2\delta x}{2} = 2x + \delta x \).
And, \( \frac{C-D}{2} = \frac{2x + 2\delta x - 2x}{2} = \frac{2\delta x}{2} = \delta x \).
Substituting these into the formula:
\( \implies \frac{dy}{dx} = \text{Lt}_{\delta x \to 0} \frac{2 \cos (2x + \delta x) \sin (\delta x)}{\delta x} \)
We can separate the limits:
\( \implies \frac{dy}{dx} = \left( \text{Lt}_{\delta x \to 0} 2 \cos (2x + \delta x) \right) \cdot \left( \text{Lt}_{\delta x \to 0} \frac{\sin (\delta x)}{\delta x} \right) \)
We know that \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \).
\( \implies \frac{dy}{dx} = 2 \cos (2x + 0) \cdot 1 \)
\( \implies \frac{dy}{dx} = 2 \cos 2x \)
In simple words: We find the derivative of sine 2x by using the basic definition. This involves breaking down the expression using a trigonometry rule and then applying a limit, which helps us find the instantaneous rate of change.

🎯 Exam Tip: When using first principles for trigonometric functions, remember to apply the sum-to-product formulas and the standard limit \( \text{Lt}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \). Careless algebra is a common source of errors.

Question 3. \( f(x) = \sqrt{3x+4}, x > -1 \)
Answer: We need to differentiate the given function.
Given \( f(x) = \sqrt{3x+4} \). We can write this as \( f(x) = (3x+4)^{1/2} \).
To differentiate both sides with respect to \( x \), we use the chain rule. This rule is very useful for functions within functions.
\( \frac{d}{dx} f(x) = f'(x) = \frac{d}{dx} (3x + 4)^{1/2} \)
Using the power rule and chain rule, which states that \( \frac{d}{dx} (g(x))^n = n(g(x))^{n-1} \cdot g'(x) \):
\( \implies f'(x) = \frac{1}{2} (3x + 4)^{1/2 - 1} \cdot \frac{d}{dx} (3x + 4) \)
\( \implies f'(x) = \frac{1}{2} (3x + 4)^{-1/2} \cdot (3 \cdot 1 + 0) \)
\( \implies f'(x) = \frac{1}{2} (3x + 4)^{-1/2} \cdot 3 \)
\( \implies f'(x) = \frac{3}{2(3x + 4)^{1/2}} \)
\( \implies f'(x) = \frac{3}{2 \sqrt{3x+4}} \)
In simple words: We find the derivative of the square root of (3x + 4) by first changing the square root into a power of 1/2. Then we use the power rule and the chain rule, which helps us differentiate functions that are inside other functions.

🎯 Exam Tip: Remember to apply the chain rule when differentiating composite functions like \( (ax+b)^n \). Differentiate the outer function first, then multiply by the derivative of the inner function.

Question 4. \( f(x) = \sqrt{4-x}, x < 4 \)
Answer: We need to find the derivative of the given function.
Given \( f(x) = \sqrt{4-x} \). This can be written as \( f(x) = (4-x)^{1/2} \).
To differentiate both sides with respect to \( x \), we apply the chain rule.
\( f'(x) = \frac{d}{dx} (4-x)^{1/2} \)
Using the power rule and chain rule:
\( \implies f'(x) = \frac{1}{2} (4-x)^{1/2 - 1} \cdot \frac{d}{dx} (4-x) \)
\( \implies f'(x) = \frac{1}{2} (4-x)^{-1/2} \cdot (0 - 1) \)
\( \implies f'(x) = \frac{1}{2} (4-x)^{-1/2} \cdot (-1) \)
\( \implies f'(x) = \frac{-1}{2(4-x)^{1/2}} \)
\( \implies f'(x) = \frac{-1}{2 \sqrt{4-x}} \)
In simple words: To differentiate the square root of (4 minus x), we convert it to (4-x) to the power of 1/2. Then we use the rule for powers and also the chain rule because there is a function (4-x) inside the power function.

🎯 Exam Tip: Be careful with the sign when differentiating \( (4-x) \). The derivative of \( -x \) is \( -1 \), which is crucial for the final answer.

Question 5. \( f(x) = \frac{3x+4}{4x+3}\left(x \neq \frac{-3}{4}\right) \)
Answer: We need to find the differential coefficient of the given rational function.
Given \( f(x) = \frac{3x+4}{4x+3} \).
To differentiate both sides with respect to \( x \), we use the quotient rule. The quotient rule is essential for functions that are fractions.
The quotient rule states: if \( f(x) = \frac{u}{v} \), then \( f'(x) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
Here, \( u = 3x+4 \) and \( v = 4x+3 \).
So, \( \frac{du}{dx} = 3 \) and \( \frac{dv}{dx} = 4 \).
Applying the quotient rule:
\( f'(x) = \frac{(4x + 3) \frac{d}{dx} (3x+4) - (3x + 4) \frac{d}{dx} (4x + 3)}{(4x + 3)^2} \)
\( \implies f'(x) = \frac{(4x + 3) (3) - (3x + 4) (4)}{(4x + 3)^2} \)
\( \implies f'(x) = \frac{12x + 9 - (12x + 16)}{(4x + 3)^2} \)
\( \implies f'(x) = \frac{12x + 9 - 12x - 16}{(4x + 3)^2} \)
\( \implies f'(x) = \frac{-7}{(4x + 3)^2} \)
In simple words: To differentiate this fraction, we use the quotient rule. This rule helps us find the derivative when one function is divided by another. We differentiate the top part, then the bottom part, and combine them using a special formula.

🎯 Exam Tip: The quotient rule involves a subtraction in the numerator, so be very careful with signs, especially when expanding the terms. Remember to square the denominator as well.

Question 6. \( f(x) = \sqrt{x^2+1} \)
Answer: We need to find the differential coefficient of the given function.
Given \( f(x) = \sqrt{x^2+1} \). This can be written as \( f(x) = (x^2+1)^{1/2} \).
To differentiate both sides with respect to \( x \), we use the chain rule. The chain rule is crucial for differentiating functions that are composed of other functions.
\( f'(x) = \frac{d}{dx} (x^2+1)^{1/2} \)
Using the power rule and chain rule:
\( \implies f'(x) = \frac{1}{2} (x^2+1)^{1/2 - 1} \cdot \frac{d}{dx} (x^2+1) \)
\( \implies f'(x) = \frac{1}{2} (x^2+1)^{-1/2} \cdot (2x + 0) \)
\( \implies f'(x) = \frac{1}{2} (x^2+1)^{-1/2} \cdot 2x \)
\( \implies f'(x) = \frac{2x}{2(x^2+1)^{1/2}} \)
\( \implies f'(x) = \frac{x}{\sqrt{x^2+1}} \)
In simple words: To find the derivative of the square root of (x squared plus 1), we first write it as (x squared plus 1) raised to the power of 1/2. Then, we use the power rule and the chain rule, which helps us differentiate functions that are nested inside one another.

🎯 Exam Tip: Always remember to differentiate the inner function when using the chain rule. For \( \sqrt{u} \), the derivative is \( \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \).

Question 7. \( (2x + 3) (x^2 - x + 2) \)
Answer: We need to find the differential coefficient of the given product of two functions.
Let \( y = (2x + 3) (x^2 - x + 2) \).
To differentiate this product, we use the product rule. This rule helps us differentiate when two functions are multiplied together.
The product rule states: if \( y = uv \), then \( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).
Here, let \( u = (2x + 3) \) and \( v = (x^2 - x + 2) \).
Then, \( \frac{du}{dx} = \frac{d}{dx} (2x + 3) = 2 \cdot 1 + 0 = 2 \).
And, \( \frac{dv}{dx} = \frac{d}{dx} (x^2 - x + 2) = 2x - 1 + 0 = 2x - 1 \).
Applying the product rule:
\( \frac{dy}{dx} = (2x + 3) (2x - 1) + (x^2 - x + 2) (2) \)
\( \implies \frac{dy}{dx} = (4x^2 - 2x + 6x - 3) + (2x^2 - 2x + 4) \)
\( \implies \frac{dy}{dx} = (4x^2 + 4x - 3) + (2x^2 - 2x + 4) \)
\( \implies \frac{dy}{dx} = 4x^2 + 4x - 3 + 2x^2 - 2x + 4 \)
Combining like terms:
\( \implies \frac{dy}{dx} = (4x^2 + 2x^2) + (4x - 2x) + (-3 + 4) \)
\( \implies \frac{dy}{dx} = 6x^2 + 2x + 1 \)
In simple words: When two functions are multiplied, we use the product rule to differentiate them. This means we differentiate one function while keeping the other the same, and then swap roles and add the results.

🎯 Exam Tip: Ensure you expand and combine terms carefully after applying the product rule. Watch out for negative signs and common algebraic errors during simplification.

Question 8. \( \tan (5x + 7) \)
Answer: We need to find the differential coefficient of the given trigonometric function.
Let \( y = \tan (5x + 7) \).
To differentiate both sides with respect to \( x \), we use the chain rule for trigonometric functions. The chain rule helps us differentiate functions that are nested inside others, like \( \tan(g(x)) \).
We know that \( \frac{d}{dx} (\tan u) = \sec^2 u \cdot \frac{du}{dx} \).
Here, \( u = 5x + 7 \).
So, \( \frac{du}{dx} = \frac{d}{dx} (5x + 7) = 5 \cdot 1 + 0 = 5 \).
Applying the chain rule:
\( \frac{dy}{dx} = \sec^2 (5x + 7) \cdot \frac{d}{dx} (5x + 7) \)
\( \implies \frac{dy}{dx} = \sec^2 (5x + 7) \cdot (5) \)
\( \implies \frac{dy}{dx} = 5 \sec^2 (5x + 7) \)
In simple words: To differentiate tangent of (5x + 7), we first find the derivative of the tangent function itself, which is secant squared. Then, because (5x + 7) is inside the tangent, we multiply by the derivative of (5x + 7) using the chain rule.

🎯 Exam Tip: When differentiating trigonometric functions using the chain rule, remember the derivative of the outer function (e.g., \( \tan u \)) and then multiply by the derivative of the inner function (e.g., \( u \)).

Question 9. \( \sin^2 (3x - 2) \)
Answer: We need to find the differential coefficient of the given function.
Let \( y = \sin^2 (3x - 2) \). This can also be written as \( y = [\sin (3x - 2)]^2 \).
To differentiate both sides with respect to \( x \), we apply the chain rule multiple times, as this is a composite function with several layers.
First, we differentiate the power function \( (u)^2 \):
\( \frac{dy}{dx} = 2 [\sin (3x - 2)]^{2-1} \cdot \frac{d}{dx} (\sin (3x - 2)) \)
\( \implies \frac{dy}{dx} = 2 \sin (3x - 2) \cdot \frac{d}{dx} (\sin (3x - 2)) \)
Next, we differentiate the sine function \( \sin (v) \), where \( v = 3x - 2 \). The derivative of \( \sin v \) is \( \cos v \cdot \frac{dv}{dx} \).
\( \implies \frac{dy}{dx} = 2 \sin (3x - 2) \cdot \cos (3x - 2) \cdot \frac{d}{dx} (3x - 2) \)
Finally, we differentiate the inner function \( (3x - 2) \):
\( \implies \frac{dy}{dx} = 2 \sin (3x - 2) \cos (3x - 2) \cdot (3) \)
Using the double angle identity \( 2 \sin A \cos A = \sin 2A \):
\( \implies \frac{dy}{dx} = 3 \cdot [2 \sin (3x - 2) \cos (3x - 2)] \)
\( \implies \frac{dy}{dx} = 3 \sin [2(3x - 2)] \)
\( \implies \frac{dy}{dx} = 3 \sin (6x - 4) \)
In simple words: To differentiate sine squared of (3x minus 2), we use the chain rule step-by-step. First, we differentiate the square part, then the sine part, and finally the (3x minus 2) part. We then use a trigonometric identity to simplify the answer.

🎯 Exam Tip: This question requires a multi-layered application of the chain rule. Differentiate from the outermost function (power) inwards (trigonometric function, then linear function) and multiply the derivatives at each step.

Question 10. \( (x^3 + \sin x)^5 \)
Answer: We need to find the differential coefficient of the given function.
Let \( y = (x^3 + \sin x)^5 \).
To differentiate both sides with respect to \( x \), we use the chain rule. The chain rule is applied when a function is raised to a power, and inside that power is another function.
The general form of the chain rule for \( [g(x)]^n \) is \( n[g(x)]^{n-1} \cdot g'(x) \).
Here, \( g(x) = x^3 + \sin x \) and \( n = 5 \).
First, differentiate the outer power function:
\( \frac{dy}{dx} = 5 (x^3 + \sin x)^{5-1} \cdot \frac{d}{dx} (x^3 + \sin x) \)
\( \implies \frac{dy}{dx} = 5 (x^3 + \sin x)^4 \cdot \frac{d}{dx} (x^3 + \sin x) \)
Next, differentiate the inner function \( (x^3 + \sin x) \):
\( \frac{d}{dx} (x^3 + \sin x) = \frac{d}{dx} (x^3) + \frac{d}{dx} (\sin x) = 3x^2 + \cos x \).
Substitute this back into the expression for \( \frac{dy}{dx} \):
\( \implies \frac{dy}{dx} = 5 (x^3 + \sin x)^4 (3x^2 + \cos x) \)
In simple words: To differentiate this expression, which is a function raised to the power of 5, we use the chain rule. We first bring the power down and reduce it by one, then multiply by the derivative of the expression inside the bracket.

🎯 Exam Tip: When using the chain rule for a power of a function, ensure you apply the power rule correctly and then remember to multiply by the derivative of the base function (the part inside the parentheses).