CBSE Class 11 Mathematics Practice Worksheet A on Differentiation, covering sum and product rules with questions on derivatives of trigonometric, polynomial, and rational functions.

Let's solve each question from the Practice Worksheet A on Differentiation (As Sum, Product) for CBSE Class 11 Mathematics.

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Very Short Answer Questions

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#### Q1. If $ y = \frac{x}{\cos x} $, find $ \frac{dy}{dx} $.

This is a quotient of two functions:
$ y = \frac{u}{v} $, where $ u = x $, $ v = \cos x $

Using Quotient Rule:
$$
\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}
$$

- $ \frac{du}{dx} = 1 $
- $ \frac{dv}{dx} = -\sin x $

So,
$$
\frac{dy}{dx} = \frac{\cos x \cdot 1 - x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos x + x \sin x}{\cos^2 x}
$$

✔ Answer: $ \boxed{\frac{\cos x + x \sin x}{\cos^2 x}} $

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#### Q2. If $ y = (1 - \cos x)(1 + \tan x) $, find $ \frac{dy}{dx} $.

Use Product Rule:
$ y = u \cdot v $, where
$ u = 1 - \cos x $, $ v = 1 + \tan x $

Then:
$$
\frac{dy}{dx} = u'v + uv'
$$

- $ u' = \sin x $
- $ v' = \sec^2 x $

So,
$$
\frac{dy}{dx} = (\sin x)(1 + \tan x) + (1 - \cos x)(\sec^2 x)
$$

✔ Answer: $ \boxed{\sin x(1 + \tan x) + (1 - \cos x)\sec^2 x} $

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#### Q3. If $ y = (x^2 - 5x + 6)(x^3 + 2) $, find $ \frac{dy}{dx} $.

Let $ u = x^2 - 5x + 6 $, $ v = x^3 + 2 $

Then:
- $ u' = 2x - 5 $
- $ v' = 3x^2 $

By product rule:
$$
\frac{dy}{dx} = u'v + uv' = (2x - 5)(x^3 + 2) + (x^2 - 5x + 6)(3x^2)
$$

Now expand:

First term:
$ (2x - 5)(x^3 + 2) = 2x^4 + 4x - 5x^3 - 10 = 2x^4 - 5x^3 + 4x - 10 $

Second term:
$ (x^2 - 5x + 6)(3x^2) = 3x^4 - 15x^3 + 18x^2 $

Add both:
$$
(2x^4 - 5x^3 + 4x - 10) + (3x^4 - 15x^3 + 18x^2) = 5x^4 - 20x^3 + 18x^2 + 4x - 10
$$

✔ Answer: $ \boxed{5x^4 - 20x^3 + 18x^2 + 4x - 10} $

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#### Q4. If $ y = (1 + 2\tan x)(5 + 4\cos x) $, find $ \frac{dy}{dx} $.

Let $ u = 1 + 2\tan x $, $ v = 5 + 4\cos x $

Then:
- $ u' = 2\sec^2 x $
- $ v' = -4\sin x $

Product rule:
$$
\frac{dy}{dx} = u'v + uv' = (2\sec^2 x)(5 + 4\cos x) + (1 + 2\tan x)(-4\sin x)
$$

Simplify:
$$
= 10\sec^2 x + 8\sec^2 x \cos x - 4\sin x - 8\tan x \sin x
$$

Note: $ \sec^2 x \cos x = \frac{1}{\cos^2 x} \cdot \cos x = \frac{1}{\cos x} = \sec x $

So:
$$
= 10\sec^2 x + 8\sec x - 4\sin x - 8\tan x \sin x
$$

Also, $ \tan x \sin x = \frac{\sin x}{\cos x} \cdot \sin x = \frac{\sin^2 x}{\cos x} $

So:
$$
\frac{dy}{dx} = 10\sec^2 x + 8\sec x - 4\sin x - 8\frac{\sin^2 x}{\cos x}
$$

But we can leave it in factored form as well.

✔ Answer: $ \boxed{2\sec^2 x(5 + 4\cos x) - 4\sin x(1 + 2\tan x)} $

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#### Q5. If $ y = \frac{ax + b}{cx + d} $, find $ \frac{dy}{dx} $.

Quotient Rule:
Let $ u = ax + b $, $ v = cx + d $

- $ u' = a $
- $ v' = c $

$$
\frac{dy}{dx} = \frac{v u' - u v'}{v^2} = \frac{(cx + d)(a) - (ax + b)(c)}{(cx + d)^2}
$$

Numerator:
$$
a(cx + d) - c(ax + b) = acx + ad - acx - bc = ad - bc
$$

So:
$$
\frac{dy}{dx} = \frac{ad - bc}{(cx + d)^2}
$$

✔ Answer: $ \boxed{\frac{ad - bc}{(cx + d)^2}} $

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Short Answer Questions

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#### Q6. Find $ \frac{dy}{dx} $ if $ y = \cot x $.

We know:
$$
\frac{d}{dx}(\cot x) = -\csc^2 x
$$

✔ Answer: $ \boxed{-\csc^2 x} $

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#### Q7. Differentiate $ \frac{x^n - a^n}{x - a} $ w.r.t. $ x $.

Let $ y = \frac{x^n - a^n}{x - a} $

Note: For $ x \ne a $, this is the definition of derivative of $ x^n $ at $ x = a $, but here we are differentiating with respect to $ x $, treating $ a $ as constant.

We can use Quotient Rule:

Let $ u = x^n - a^n $, $ v = x - a $

Then:
- $ u' = n x^{n-1} $
- $ v' = 1 $

So:
$$
\frac{dy}{dx} = \frac{(x - a)(n x^{n-1}) - (x^n - a^n)(1)}{(x - a)^2}
$$

Simplify numerator:
$$
n x^{n-1}(x - a) - (x^n - a^n) = n x^n - n a x^{n-1} - x^n + a^n = (n - 1)x^n - n a x^{n-1} + a^n
$$

Alternatively, you can factor or recognize that:
$$
\frac{x^n - a^n}{x - a} = x^{n-1} + a x^{n-2} + a^2 x^{n-3} + \cdots + a^{n-1}
$$

Then derivative:
$$
(n-1)x^{n-2} + 2a x^{n-3} + 3a^2 x^{n-4} + \cdots + (n-1)a^{n-2}
$$

But unless specified, the quotient rule form is acceptable.

✔ Answer: $ \boxed{\frac{n x^{n-1}(x - a) - (x^n - a^n)}{(x - a)^2}} $

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#### Q8. Find $ \frac{dy}{dx} $ if $ y = 1 - \frac{1}{x^2} $

Rewrite: $ y = 1 - x^{-2} $

Differentiate:
$$
\frac{dy}{dx} = 0 - (-2)x^{-3} = 2x^{-3} = \frac{2}{x^3}
$$

✔ Answer: $ \boxed{\frac{2}{x^3}} $

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#### Q9. If $ y = \frac{x}{1 + \tan x} $, find $ \frac{dy}{dx} $.

Quotient Rule: $ u = x $, $ v = 1 + \tan x $

- $ u' = 1 $
- $ v' = \sec^2 x $

$$
\frac{dy}{dx} = \frac{(1 + \tan x)(1) - x(\sec^2 x)}{(1 + \tan x)^2} = \frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2}
$$

✔ Answer: $ \boxed{\frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2}} $

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#### Q10. If $ y = \frac{x}{1 - \cot x} $, find $ \frac{dy}{dx} $.

Let $ u = x $, $ v = 1 - \cot x $

- $ u' = 1 $
- $ v' = \csc^2 x $ (since $ \frac{d}{dx}(\cot x) = -\csc^2 x $)

So:
$$
\frac{dy}{dx} = \frac{(1 - \cot x)(1) - x(\csc^2 x)}{(1 - \cot x)^2} = \frac{1 - \cot x - x \csc^2 x}{(1 - \cot x)^2}
$$

✔ Answer: $ \boxed{\frac{1 - \cot x - x \csc^2 x}{(1 - \cot x)^2}} $

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✔ Final Answers Summary:

| Q | Answer |
|---|--------|
| Q1 | $ \frac{\cos x + x \sin x}{\cos^2 x} $ |
| Q2 | $ \sin x(1 + \tan x) + (1 - \cos x)\sec^2 x $ |
| Q3 | $ 5x^4 - 20x^3 + 18x^2 + 4x - 10 $ |
| Q4 | $ 2\sec^2 x(5 + 4\cos x) - 4\sin x(1 + 2\tan x) $ |
| Q5 | $ \frac{ad - bc}{(cx + d)^2} $ |
| Q6 | $ -\csc^2 x $ |
| Q7 | $ \frac{n x^{n-1}(x - a) - (x^n - a^n)}{(x - a)^2} $ |
| Q8 | $ \frac{2}{x^3} $ |
| Q9 | $ \frac{1 + \tan x - x \sec^2 x}{(1 + \tan x)^2} $ |
| Q10 | $ \frac{1 - \cot x - x \csc^2 x}{(1 - \cot x)^2} $ |

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