Samacheer Kalvi Class 11 Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.2

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability And Methods Of Differentiation Ex 10.2

Find the derivatives of the following functions with respect to corresponding independent variables:

Question 1. f(x) = x - 3 sin x
Answer:
Given the function: \( f(x) = x - 3 \sin x \)
To find the derivative, we differentiate each term with respect to \(x\).
The derivative of \(x\) is \(1\).
The derivative of \( \sin x \) is \( \cos x \).
So, \( f'(x) = \frac{d}{dx}(x) - 3 \frac{d}{dx}(\sin x) \)
\( f'(x) = 1 - 3 (\cos x) \)
\( f'(x) = 1 - 3 \cos x \)
The derivative helps us find the rate of change of a function.
In simple words: To find the derivative, we differentiate \(x\) to get 1 and \( \sin x \) to get \( \cos x \), then put them together.

🎯 Exam Tip: Remember to apply differentiation rules to each term separately in a sum or difference of functions.

Question 2. y = sin x + cos x
Answer:
Given the function: \( y = \sin x + \cos x \)
To find the derivative \( \frac{dy}{dx} \), we differentiate each term with respect to \(x\).
The derivative of \( \sin x \) is \( \cos x \).
The derivative of \( \cos x \) is \( -\sin x \).
So, \( \frac{dy}{dx} = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\cos x) \)
\( \frac{dy}{dx} = \cos x - \sin x \)
The derivative of sine is cosine, and the derivative of cosine is negative sine.
In simple words: We find the derivative of \( \sin x \) which is \( \cos x \), and the derivative of \( \cos x \) which is \( -\sin x \). Then we combine them.

🎯 Exam Tip: Be careful with the signs when differentiating trigonometric functions; the derivative of \( \cos x \) is negative.

Question 3. f(x) = x sin x
Answer:
Given the function: \( f(x) = x \sin x \)
This is a product of two functions, \( u = x \) and \( v = \sin x \). We use the product rule: \( (uv)' = u'v + uv' \).
Let \( u = x \)
Then, \( u' = \frac{du}{dx} = 1 \)
Let \( v = \sin x \)
Then, \( v' = \frac{dv}{dx} = \cos x \)
Applying the product rule:
\( f'(x) = (1)(\sin x) + (x)(\cos x) \)
\( f'(x) = \sin x + x \cos x \)
This rule, called the product rule, helps us find derivatives of functions multiplied together.
In simple words: When two functions are multiplied, like \(x\) and \( \sin x \), we use a special rule. We take the derivative of the first, multiply by the second, then add it to the first multiplied by the derivative of the second.

🎯 Exam Tip: Remember the product rule formula: \( (uv)' = u'v + uv' \) and apply it carefully, identifying \(u\), \(v\), \(u'\), and \(v'\) correctly.

Question 4. y = cos x - 2 tan x
Answer:
Given the function: \( y = \cos x - 2 \tan x \)
To find the derivative \( \frac{dy}{dx} \), we differentiate each term with respect to \(x\).
The derivative of \( \cos x \) is \( -\sin x \).
The derivative of \( \tan x \) is \( \sec² x \).
So, \( \frac{dy}{dx} = \frac{d}{dx}(\cos x) - 2 \frac{d}{dx}(\tan x) \)
\( \frac{dy}{dx} = -\sin x - 2 \sec² x \)
The derivative of \( \cos x \) is \( -\sin x \), and the derivative of \( \tan x \) is \( \sec² x \).
In simple words: We find the derivative of \( \cos x \) which is \( -\sin x \), and the derivative of \( \tan x \) which is \( \sec² x \), then combine them with the constant.

🎯 Exam Tip: Do not forget the constant multiplier (like 2 here) when differentiating terms, and it remains in the result.

Question 5. g(t) = t³ cos t
Answer:
Given the function: \( g(t) = t³ \cos t \)
This is a product of two functions, \( u = t³ \) and \( v = \cos t \). We use the product rule: \( (uv)' = u'v + uv' \).
Let \( u = t³ \)
Then, \( u' = \frac{du}{dt} = 3t² \)
Let \( v = \cos t \)
Then, \( v' = \frac{dv}{dt} = -\sin t \)
Applying the product rule:
\( g'(t) = (3t²)(\cos t) + (t³)(-\sin t) \)
\( g'(t) = 3t² \cos t - t³ \sin t \)
The product rule is useful when finding the derivative of a function made by multiplying two other functions.
In simple words: We use the product rule because \( g(t) \) is made of two parts, \( t³ \) and \( \cos t \), multiplied together. We find the derivative by following the product rule formula.

🎯 Exam Tip: Pay attention to the variable of differentiation (here, \(t\)) and ensure all derivatives are taken with respect to that variable.

Question 6. g(t) = 4 sec t + tan t
Answer:
Given the function: \( g(t) = 4 \sec t + \tan t \)
To find the derivative \( g'(t) \), we differentiate each term with respect to \(t\).
The derivative of \( \sec t \) is \( \sec t \tan t \).
The derivative of \( \tan t \) is \( \sec² t \).
So, \( g'(t) = 4 \frac{d}{dt}(\sec t) + \frac{d}{dt}(\tan t) \)
\( g'(t) = 4 (\sec t \tan t) + \sec² t \)
\( g'(t) = 4 \sec t \tan t + \sec² t \)
The derivative of \( \sec t \) is \( \sec t \tan t \), and the derivative of \( \tan t \) is \( \sec² t \).
In simple words: We differentiate each part of the function separately. \( \sec t \) becomes \( \sec t \tan t \), and \( \tan t \) becomes \( \sec² t \).

🎯 Exam Tip: Memorize the derivatives of common trigonometric functions, as they are frequently used in calculus problems.

Question 7. y = e^x sin x
Answer:
Given the function: \( y = e^x \sin x \)
This is a product of two functions, \( u = e^x \) and \( v = \sin x \). We use the product rule: \( (uv)' = u'v + uv' \).
Let \( u = e^x \)
Then, \( u' = \frac{du}{dx} = e^x \)
Let \( v = \sin x \)
Then, \( v' = \frac{dv}{dx} = \cos x \)
Applying the product rule:
\( \frac{dy}{dx} = (e^x)(\sin x) + (e^x)(\cos x) \)
Factor out \( e^x \):
\( \frac{dy}{dx} = e^x (\sin x + \cos x) \)
The exponential function \( e^x \) is unique because its derivative is also \( e^x \).
In simple words: We use the product rule for \( e^x \) and \( \sin x \). The derivative of \( e^x \) is itself, and the derivative of \( \sin x \) is \( \cos x \).

🎯 Exam Tip: The derivative of \( e^x \) is one of the easiest to remember, as it remains \( e^x \).

Question 8. y = \( \frac{\tan x}{x} \)
Answer:
Given the function: \( y = \frac{\tan x}{x} \)
This is a quotient of two functions, \( u = \tan x \) and \( v = x \). We use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
Let \( u = \tan x \)
Then, \( u' = \frac{du}{dx} = \sec² x \)
Let \( v = x \)
Then, \( v' = \frac{dv}{dx} = 1 \)
Applying the quotient rule:
\( \frac{dy}{dx} = \frac{(\sec² x)(x) - (\tan x)(1)}{x²} \)
\( \frac{dy}{dx} = \frac{x \sec² x - \tan x}{x²} \)
The quotient rule is used to find the derivative of a fraction where both the top and bottom are functions of \(x\).
In simple words: When a function is divided by another, we use the quotient rule. We differentiate the top, multiply by the bottom, subtract the top multiplied by the derivative of the bottom, all divided by the bottom squared.

🎯 Exam Tip: Be careful with the order of terms in the numerator of the quotient rule; it's \( u'v - uv' \), not the other way around.

Question 9. y = \( \frac{\sin x}{1+\cos x} \)
Answer:
Given the function: \( y = \frac{\sin x}{1+\cos x} \)
This is a quotient of two functions, \( u = \sin x \) and \( v = 1 + \cos x \). We use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
Let \( u = \sin x \)
Then, \( u' = \frac{du}{dx} = \cos x \)
Let \( v = 1 + \cos x \)
Then, \( v' = \frac{dv}{dx} = 0 - \sin x = -\sin x \)
Applying the quotient rule:
\( \frac{dy}{dx} = \frac{(\cos x)(1 + \cos x) - (\sin x)(-\sin x)}{(1 + \cos x)²} \)
\( \frac{dy}{dx} = \frac{\cos x + \cos² x + \sin² x}{(1 + \cos x)²} \)
Using the trigonometric identity \( \cos² x + \sin² x = 1 \):
\( \frac{dy}{dx} = \frac{\cos x + 1}{(1 + \cos x)²} \)
\( \frac{dy}{dx} = \frac{1}{1 + \cos x} \)
Remember the identity \( \sin² x + \cos² x = 1 \) which simplifies the expression significantly.
In simple words: We apply the division rule for derivatives. After doing the math, we use the fact that \( \sin² x + \cos² x \) equals 1 to make the answer simpler.

🎯 Exam Tip: Always look for trigonometric identities that can simplify the expression after differentiation, especially for \( \sin² x + \cos² x \).

Question 10. y = \( \frac{x}{\sin x+\cos x} \)
Answer:
Given the function: \( y = \frac{x}{\sin x+\cos x} \)
This is a quotient of two functions, \( u = x \) and \( v = \sin x + \cos x \). We use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
Let \( u = x \)
Then, \( u' = \frac{du}{dx} = 1 \)
Let \( v = \sin x + \cos x \)
Then, \( v' = \frac{dv}{dx} = \cos x - \sin x \)
Applying the quotient rule:
\( \frac{dy}{dx} = \frac{(1)(\sin x + \cos x) - (x)(\cos x - \sin x)}{(\sin x + \cos x)²} \)
\( \frac{dy}{dx} = \frac{\sin x + \cos x - x \cos x + x \sin x}{(\sin x + \cos x)²} \)
Rearranging terms in the numerator:
\( \frac{dy}{dx} = \frac{(1 + x) \sin x + (1 - x) \cos x}{(\sin x + \cos x)²} \)
This problem involves applying the quotient rule and then carefully simplifying the numerator by grouping terms.
In simple words: We use the division rule for derivatives. We differentiate the top part, multiply by the bottom, subtract the top multiplied by the derivative of the bottom, all divided by the bottom part squared. Then we combine similar terms on top.

🎯 Exam Tip: When simplifying the numerator, group terms involving \( \sin x \) and \( \cos x \) to present the answer neatly.

Question 11. y = \( \frac{\tan x-1}{\sec x} \)
Answer:
Given the function: \( y = \frac{\tan x-1}{\sec x} \)
This is a quotient of two functions. We use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
Let \( u = \tan x - 1 \)
Then, \( u' = \frac{du}{dx} = \sec² x - 0 = \sec² x \)
Let \( v = \sec x \)
Then, \( v' = \frac{dv}{dx} = \sec x \tan x \)
Applying the quotient rule:
\( \frac{dy}{dx} = \frac{(\sec² x)(\sec x) - (\tan x - 1)(\sec x \tan x)}{(\sec x)²} \)
Factor out \( \sec x \) from the numerator:
\( \frac{dy}{dx} = \frac{\sec x [\sec² x - (\tan x - 1)\tan x]}{(\sec x)²} \)
\( \frac{dy}{dx} = \frac{\sec² x - \tan² x + \tan x}{\sec x} \)
Using the trigonometric identity \( \sec² x - \tan² x = 1 \):
\( \frac{dy}{dx} = \frac{1 + \tan x}{\sec x} \)
This can be further simplified:
\( \frac{dy}{dx} = \frac{1}{\sec x} + \frac{\tan x}{\sec x} \)
\( \frac{dy}{dx} = \cos x + \frac{\sin x / \cos x}{1 / \cos x} \)
\( \frac{dy}{dx} = \cos x + \sin x \)
Simplifying the function first using trigonometric identities can sometimes make differentiation easier.
In simple words: We use the division rule for derivatives. After the calculation, we use the identity \( \sec² x - \tan² x = 1 \) to simplify the answer. Then we convert to sine and cosine for the final simple form.

🎯 Exam Tip: Sometimes, simplifying the original function using trigonometric identities *before* differentiating can save a lot of work.

Question 12. y = \( \frac{\sin x}{x^{2}} \)
Answer:
Given the function: \( y = \frac{\sin x}{x^{2}} \)
This is a quotient of two functions, \( u = \sin x \) and \( v = x² \). We use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
Let \( u = \sin x \)
Then, \( u' = \frac{du}{dx} = \cos x \)
Let \( v = x² \)
Then, \( v' = \frac{dv}{dx} = 2x \)
Applying the quotient rule:
\( \frac{dy}{dx} = \frac{(\cos x)(x²) - (\sin x)(2x)}{(x²)²} \)
\( \frac{dy}{dx} = \frac{x² \cos x - 2x \sin x}{x^4} \)
Factor out \( x \) from the numerator:
\( \frac{dy}{dx} = \frac{x(x \cos x - 2 \sin x)}{x^4} \)
Cancel out one \( x \) term:
\( \frac{dy}{dx} = \frac{x \cos x - 2 \sin x}{x³} \)
When simplifying fractions with derivatives, always look for common factors in the numerator and denominator to cancel out.
In simple words: We use the division rule for derivatives. After the steps, we simplify the fraction by taking out a common \(x\) from the top and canceling it with one \(x\) from the bottom.

🎯 Exam Tip: Always simplify the final expression by canceling common factors in the numerator and denominator to get the most reduced form.

Question 13. y = tan θ (sin θ + cos θ)
Answer:
Given the function: \( y = \tan \theta (\sin \theta + \cos \theta) \)
This is a product of two functions, \( u = \tan \theta \) and \( v = \sin \theta + \cos \theta \). We use the product rule: \( (uv)' = u'v + uv' \).
Let \( u = \tan \theta \)
Then, \( u' = \frac{du}{d\theta} = \sec² \theta \)
Let \( v = \sin \theta + \cos \theta \)
Then, \( v' = \frac{dv}{d\theta} = \cos \theta - \sin \theta \)
Applying the product rule:
\( \frac{dy}{d\theta} = (\sec² \theta)(\sin \theta + \cos \theta) + (\tan \theta)(\cos \theta - \sin \theta) \)
Expand the terms:
\( \frac{dy}{d\theta} = \sec² \theta \sin \theta + \sec² \theta \cos \theta + \tan \theta \cos \theta - \tan \theta \sin \theta \)
Now, express in terms of \( \sin \theta \) and \( \cos \theta \):
\( \frac{dy}{d\theta} = \frac{1}{\cos² \theta} \sin \theta + \frac{1}{\cos² \theta} \cos \theta + \frac{\sin \theta}{\cos \theta} \cos \theta - \frac{\sin \theta}{\cos \theta} \sin \theta \)
\( \frac{dy}{d\theta} = \frac{\sin \theta}{\cos² \theta} + \frac{1}{\cos \theta} + \sin \theta - \frac{\sin² \theta}{\cos \theta} \)
Rearranging and grouping terms:
\( \frac{dy}{d\theta} = (\sin \theta - \frac{\sin² \theta}{\cos \theta}) + \frac{\sin \theta}{\cos² \theta} + \frac{1}{\cos \theta} \)
\( \frac{dy}{d\theta} = \sin \theta (1 - \frac{\sin \theta}{\cos \theta}) + \sin \theta \sec² \theta + \sec \theta \)
\( \frac{dy}{d\theta} = \sin \theta (1 - \tan \theta) + \sin \theta \sec² \theta + \sec \theta \)
Alternatively, from \( \frac{dy}{d\theta} = \sec² \theta \sin \theta + \sec² \theta \cos \theta + \tan \theta \cos \theta - \tan \theta \sin \theta \)
\( \frac{dy}{d\theta} = \frac{\sin \theta}{\cos² \theta} + \frac{1}{\cos \theta} + \sin \theta - \frac{\sin² \theta}{\cos \theta} \)
Combine \( \frac{1}{\cos \theta} \) and \( \sin \theta \):
\( \frac{dy}{d\theta} = \sin \theta \sec² \theta + \sec \theta + \sin \theta - \tan \theta \sin \theta \)
Another way to group: \( \sin \theta + \cos \theta + \sec \theta \tan \theta \)
From the source's simplification:
\( \frac{dy}{d\theta} = \sin \theta - \sin² \theta \sec \theta + \tan \theta \sec \theta + \sec \theta \)
\( \frac{dy}{d\theta} = \sin \theta + (1 - \sin² \theta) \sec \theta + \sec \theta \tan \theta \)
\( \frac{dy}{d\theta} = \sin \theta + \cos² \theta \frac{1}{\cos \theta} + \sec \theta \tan \theta \)
\( \frac{dy}{d\theta} = \sin \theta + \cos \theta + \sec \theta \tan \theta \)
Sometimes, expanding the original function first and then differentiating each term can be an alternative approach to the product rule.
In simple words: We use the product rule to find the derivative of this function. Then, we simplify the result by using trigonometric identities to get a cleaner final expression.

🎯 Exam Tip: After applying the product rule, use fundamental trigonometric identities (like \( \sec^2 \theta - \tan^2 \theta = 1 \) or \( \sin^2 \theta + \cos^2 \theta = 1 \)) to simplify the answer.

Question 14. y = cosec x . cot x
Answer:
Given the function: \( y = \text{cosec } x . \text{cot } x \)
This is a product of two functions, \( u = \text{cosec } x \) and \( v = \text{cot } x \). We use the product rule: \( (uv)' = u'v + uv' \).
Let \( u = \text{cosec } x \)
Then, \( u' = \frac{du}{dx} = -\text{cosec } x \text{cot } x \)
Let \( v = \text{cot } x \)
Then, \( v' = \frac{dv}{dx} = -\text{cosec² } x \)
Applying the product rule:
\( \frac{dy}{dx} = (-\text{cosec } x \text{cot } x)(\text{cot } x) + (\text{cosec } x)(-\text{cosec² } x) \)
\( \frac{dy}{dx} = -\text{cosec } x \text{cot² } x - \text{cosec³ } x \)
Factor out \( -\text{cosec } x \):
\( \frac{dy}{dx} = -\text{cosec } x (\text{cot² } x + \text{cosec² } x) \)
To express in terms of \( \sin x \) and \( \cos x \):
\( \frac{dy}{dx} = -\frac{1}{\sin x} \left(\frac{\cos² x}{\sin² x} + \frac{1}{\sin² x}\right) \)
\( \frac{dy}{dx} = -\frac{1}{\sin x} \left(\frac{\cos² x + 1}{\sin² x}\right) \)
\( \frac{dy}{dx} = -\frac{1 + \cos² x}{\sin³ x} \)
Remembering the derivatives of reciprocal trigonometric functions like \( \text{cosec } x \) and \( \text{cot } x \) is key here.
In simple words: We use the product rule for \( \text{cosec } x \) and \( \text{cot } x \). After differentiating, we combine the terms and express them using sine and cosine for the final answer.

🎯 Exam Tip: Be mindful of the negative signs that often appear when differentiating inverse trigonometric functions.

Question 15. y = x sin x cos x
Answer:
Given the function: \( y = x \sin x \cos x \)
We can simplify the function using the identity \( 2 \sin x \cos x = \sin(2x) \).
So, \( y = x \left(\frac{1}{2} \sin(2x)\right) \)
\( y = \frac{1}{2} x \sin(2x) \)
Now, we use the product rule for \( u = \frac{1}{2} x \) and \( v = \sin(2x) \).
Let \( u = \frac{1}{2} x \)
Then, \( u' = \frac{du}{dx} = \frac{1}{2} \)
Let \( v = \sin(2x) \)
Then, \( v' = \frac{dv}{dx} = \cos(2x) \cdot 2 = 2 \cos(2x) \) (using the chain rule)
Applying the product rule:
\( \frac{dy}{dx} = u'v + uv' \)
\( \frac{dy}{dx} = \left(\frac{1}{2}\right) \sin(2x) + \left(\frac{1}{2} x\right) (2 \cos(2x)) \)
\( \frac{dy}{dx} = \frac{1}{2} \sin(2x) + x \cos(2x) \)
We can also write \( \frac{1}{2} \sin(2x) \) as \( \sin x \cos x \):
\( \frac{dy}{dx} = \sin x \cos x + x \cos(2x) \)
Recognizing and using trigonometric identities like \( 2 \sin x \cos x = \sin(2x) \) can simplify complex differentiation problems.
In simple words: First, we use a trigonometry rule to make the function simpler. Then, we use the product rule to find the derivative of the new, simpler function.

🎯 Exam Tip: Look for opportunities to simplify the original function using trigonometric identities before differentiating, as it can make the process much easier.

Question 16. y = \( e^{-x} . \log x \)
Answer:
Given the function: \( y = e^{-x} \log x \)
This is a product of two functions, \( u = e^{-x} \) and \( v = \log x \). We use the product rule: \( (uv)' = u'v + uv' \).
Let \( u = e^{-x} \)
Then, \( u' = \frac{du}{dx} = e^{-x} (-1) = -e^{-x} \) (using the chain rule)
Let \( v = \log x \)
Then, \( v' = \frac{dv}{dx} = \frac{1}{x} \)
Applying the product rule:
\( \frac{dy}{dx} = (-e^{-x})(\log x) + (e^{-x})\left(\frac{1}{x}\right) \)
\( \frac{dy}{dx} = -e^{-x} \log x + \frac{e^{-x}}{x} \)
Factor out \( e^{-x} \):
\( \frac{dy}{dx} = e^{-x} \left(\frac{1}{x} - \log x\right) \)
The chain rule is implicitly used when differentiating \( e^{-x} \) to \( e^{-x} \cdot (-1) \).
In simple words: We use the product rule for the two parts of the function. Remember that the derivative of \( e^{-x} \) involves multiplying by -1. Then we combine the terms and factor out \( e^{-x} \).

🎯 Exam Tip: Do not forget the chain rule when differentiating composite exponential functions like \( e^{-x} \).

Question 17. y = \( (x² + 5) \log (1 + x) e^{-3x} \)
Answer:
Given the function: \( y = (x² + 5) \log (1 + x) e^{-3x} \)
This is a product of three functions: \( f_1 = (x² + 5) \), \( f_2 = \log (1 + x) \), and \( f_3 = e^{-3x} \).
The product rule for three functions is \( (f_1 f_2 f_3)' = f_1' f_2 f_3 + f_1 f_2' f_3 + f_1 f_2 f_3' \).
First, find the derivatives of each individual function:
\( f_1' = \frac{d}{dx}(x² + 5) = 2x \)
\( f_2' = \frac{d}{dx}(\log (1 + x)) = \frac{1}{1 + x} \) (using the chain rule)
\( f_3' = \frac{d}{dx}(e^{-3x}) = e^{-3x} (-3) = -3e^{-3x} \) (using the chain rule)
Now, apply the product rule:
\( \frac{dy}{dx} = (2x)(\log (1 + x))(e^{-3x}) + (x² + 5)\left(\frac{1}{1 + x}\right)(e^{-3x}) + (x² + 5)(\log (1 + x))(-3e^{-3x}) \)
Factor out \( e^{-3x} \) from all terms:
\( \frac{dy}{dx} = e^{-3x} \left[ 2x \log (1 + x) + \frac{x² + 5}{1 + x} - 3(x² + 5) \log (1 + x) \right] \)
Group the terms containing \( \log (1 + x) \):
\( \frac{dy}{dx} = e^{-3x} \left[ \frac{x² + 5}{1 + x} + (2x - 3(x² + 5)) \log (1 + x) \right] \)
The product rule can be extended to multiple functions by taking the derivative of each function one at a time while keeping others constant.
In simple words: We use a special product rule for three functions. We find the derivative of each part, then combine them according to the rule, and finally group similar terms and factor out the common exponential part.

🎯 Exam Tip: For products of three or more functions, methodically apply the extended product rule by differentiating one function at a time while keeping the others unchanged.

Question 18. y = sin x°
Answer:
Given the function: \( y = \sin x° \)
Before differentiating, convert the angle from degrees to radians, as calculus functions operate on radians.
We know that \( 1° = \frac{\pi}{180} \) radians. So, \( x° = x \frac{\pi}{180} \) radians.
Substitute this into the function:
\( y = \sin \left(x \frac{\pi}{180}\right) \)
Now, differentiate with respect to \(x\) using the chain rule:
\( \frac{dy}{dx} = \cos \left(x \frac{\pi}{180}\right) \cdot \frac{d}{dx}\left(x \frac{\pi}{180}\right) \)
\( \frac{dy}{dx} = \cos \left(x \frac{\pi}{180}\right) \cdot \frac{\pi}{180} \)
Rewrite \( x \frac{\pi}{180} \) back as \( x° \):
\( \frac{dy}{dx} = \frac{\pi}{180} \cos x° \)
Calculus functions like sine and cosine are defined for radian measure, so angles in degrees must always be converted first.
In simple words: First, we change the angle from degrees to radians because calculus rules work with radians. Then, we find the derivative of sine using the chain rule, which includes multiplying by the constant factor from the radian conversion.

🎯 Exam Tip: Always convert angles from degrees to radians when working with trigonometric functions in calculus problems to ensure correct derivatives.

Question 19. y = \( \log_{10} x \)
Answer:
Given the function: \( y = \log_{10} x \)
To differentiate a logarithm with a base other than \(e\), we use the change of base formula to convert it to a natural logarithm:
\( \log_b a = \frac{\ln a}{\ln b} \)
So, \( \log_{10} x = \frac{\ln x}{\ln 10} \)
Now, differentiate with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\ln x}{\ln 10}\right) \)
Since \( \frac{1}{\ln 10} \) is a constant, we can take it out of the differentiation:
\( \frac{dy}{dx} = \frac{1}{\ln 10} \frac{d}{dx}(\ln x) \)
The derivative of \( \ln x \) is \( \frac{1}{x} \):
\( \frac{dy}{dx} = \frac{1}{\ln 10} \cdot \frac{1}{x} \)
\( \frac{dy}{dx} = \frac{1}{x \ln 10} \)
The change of base formula for logarithms is crucial here to express the logarithm in a base suitable for differentiation (usually base \(e\)).
In simple words: We change the logarithm from base 10 to a natural logarithm (base \(e\)) using a special formula. Since \( \ln 10 \) is just a number, we then find the derivative of \( \ln x \), which is \( 1/x \).

🎯 Exam Tip: Remember the change of base formula for logarithms and that only logarithms with base \(e\) (natural logarithms) have a simple derivative of \( 1/x \).

Question 20. Draw the function f'(x) if f(x) = 2x² – 5x + 3
Answer:
First, find the derivative of the given function \( f(x) \):
\( f(x) = 2x² - 5x + 3 \)
Differentiate each term with respect to \(x\):
\( f'(x) = \frac{d}{dx}(2x²) - \frac{d}{dx}(5x) + \frac{d}{dx}(3) \)
\( f'(x) = 2(2x) - 5(1) + 0 \)
\( f'(x) = 4x - 5 \)
This is a linear function, which means its graph is a straight line. Let \( y = f'(x) \), so \( y = 4x - 5 \).
To draw the function, find two points on the line:
If \( x = 0 \), then \( y = 4(0) - 5 = -5 \). So, point is \( (0, -5) \).
If \( y = 0 \), then \( 0 = 4x - 5 \implies 4x = 5 \implies x = \frac{5}{4} = 1.25 \). So, point is \( (1.25, 0) \).
Plot these points and draw a straight line through them.
The derivative of a quadratic function is always a linear function, representing the slope of the quadratic at any point.
In simple words: First, we find the derivative of \( f(x) \), which is \( f'(x) = 4x - 5 \). This is a straight line equation. To draw it, we find where the line crosses the X and Y axes, then connect those points.

🎯 Exam Tip: To draw a linear function, find its x-intercept (set y=0) and y-intercept (set x=0) and then connect these two points with a straight line.