Class 11-NCERT Solutions-Chapter-12-Limits and Derivatives

Answer:

\( = 3 + 3 = 6 \)

Answer:

\( = \pi - \frac{22}{7} \)

Answer:

\( = \pi (1)^2 = \pi \)

Answer:

\( = \frac{4(4) + 3}{4 - 2} = \frac{16 + 3}{2} = \frac{19}{2} \)

Answer:

\( = \frac{(-1)^{10} + (-1)^5 + 1}{-1 - 1} = \frac{1 - 1 + 1}{-2} = -\frac{1}{2} \)

Answer:

Put \( y = x+1 \). As \( x \to 0, y \to 1 \).

Limit becomes \( \lim_{y \to 1} \frac{y^5 - 1^5}{y - 1} \)

Using formula \( \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1} \):

\( = 5(1)^{5-1} = 5(1) = 5 \)

Answer:

At \( x=2 \), form is \( \frac{0}{0} \). Factorize:

\( \lim_{x \to 2} \frac{(3x+5)(x-2)}{(x-2)(x+2)} = \lim_{x \to 2} \frac{3x+5}{x+2} \)

\( = \frac{3(2)+5}{2+2} = \frac{11}{4} \)

Answer:

At \( x=3 \), form is \( \frac{0}{0} \). Factorize:

Numerator: \( x^4 - 3^4 = (x^2-9)(x^2+9) = (x-3)(x+3)(x^2+9) \)

Denominator: \( 2x^2-6x+x-3 = 2x(x-3)+1(x-3) = (2x+1)(x-3) \)

\( \lim_{x \to 3} \frac{(x-3)(x+3)(x^2+9)}{(2x+1)(x-3)} = \frac{(3+3)(9+9)}{2(3)+1} = \frac{6 \times 18}{7} = \frac{108}{7} \)

Answer:

\( = \frac{a(0)+b}{c(0)+1} = \frac{b}{1} = b \)

Answer:

Let \( z^{\frac{1}{6}} = x \), then \( z \to 1 \Rightarrow x \to 1 \).

\( z^{\frac{1}{3}} = (x^6)^{\frac{1}{3}} = x^2 \).

Limit becomes \( \lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} (x+1) = 2 \).

Answer:

\( = \frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a} = \frac{a+b+c}{c+b+a} = 1 \)

Answer:

Simplify numerator: \( \frac{2+x}{2x} \).

\( \lim_{x \to -2} \frac{x+2}{2x(x+2)} = \lim_{x \to -2} \frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4} \)

Answer:

\( = \lim_{x \to 0} \frac{\sin ax}{ax} \times \frac{a}{b} \)

Using \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \):

\( = 1 \times \frac{a}{b} = \frac{a}{b} \)

Answer:

Divide numerator and denominator by \( x \):

\( = \frac{\lim_{x \to 0} \frac{\sin ax}{x}}{\lim_{x \to 0} \frac{\sin bx}{x}} = \frac{a \lim_{ax \to 0} \frac{\sin ax}{ax}}{b \lim_{bx \to 0} \frac{\sin bx}{bx}} = \frac{a(1)}{b(1)} = \frac{a}{b} \)

Answer:

Put \( y = \pi - x \). As \( x \to \pi, y \to 0 \).

Limit becomes: \( \lim_{y \to 0} \frac{\sin y}{\pi y} = \frac{1}{\pi} \lim_{y \to 0} \frac{\sin y}{y} = \frac{1}{\pi} \times 1 = \frac{1}{\pi} \)

Answer:

\( = \frac{\cos 0}{\pi - 0} = \frac{1}{\pi} \)

Answer:

Using \( 1 - \cos \theta = 2\sin^2(\frac{\theta}{2}) \):

\( = \lim_{x \to 0} \frac{-(1 - \cos 2x)}{-(1 - \cos x)} = \lim_{x \to 0} \frac{2\sin^2 x}{2\sin^2(x/2)} \)

\( = \lim_{x \to 0} \frac{(\frac{\sin x}{x})^2 \cdot x^2}{(\frac{\sin(x/2)}{x/2})^2 \cdot (x/2)^2} = \frac{1^2 \cdot x^2}{1^2 \cdot \frac{x^2}{4}} = 4 \)

Answer:

Factor out x in numerator: \( \lim_{x \to 0} \frac{x(a + \cos x)}{b \sin x} \)

\( = \frac{1}{b} \lim_{x \to 0} \frac{x}{\sin x} \cdot (a + \cos x) = \frac{1}{b} \cdot 1 \cdot (a + \cos 0) = \frac{a+1}{b} \)

Answer:

\( = 0 \cdot \sec 0 = 0 \cdot 1 = 0 \)

Answer:

Divide numerator and denominator by x:

\( = \frac{\lim (\frac{\sin ax}{x} + b)}{\lim (a + \frac{\sin bx}{x})} = \frac{a(1) + b}{a + b(1)} = \frac{a+b}{a+b} = 1 \)

Answer:

\( = \lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right) = \lim_{x \to 0} \frac{1 - \cos x}{\sin x} \)

\( = \lim_{x \to 0} \frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)} = \lim_{x \to 0} \tan(x/2) = \tan 0 = 0 \)

Answer:

Let \( h = x - \frac{\pi}{2} \Rightarrow x = \frac{\pi}{2} + h \). As \( x \to \frac{\pi}{2}, h \to 0 \).

Numerator: \( \tan 2(\frac{\pi}{2} + h) = \tan(\pi + 2h) = \tan 2h \).

Limit: \( \lim_{h \to 0} \frac{\tan 2h}{h} = 2 \).

Answer:

For \( \lim_{x \to 0} f(x) \):

LHL \( = \lim_{x \to 0^-} (2x+3) = 3 \)

RHL \( = \lim_{x \to 0^+} 3(x+1) = 3(1) = 3 \)

Since LHL = RHL, limit is 3.

For \( \lim_{x \to 1} f(x) \):

Function is defined as \( 3(x+1) \) around \( x=1 \).

\( \lim_{x \to 1} 3(x+1) = 3(1+1) = 6 \)

Answer:

LHL \( = \lim_{x \to 1^-} (x^2-1) = 1-1 = 0 \)

RHL \( = \lim_{x \to 1^+} (-x^2-1) = -1-1 = -2 \)

Since LHL \( \ne \) RHL, limit does not exist.

Answer:

LHL \( = \lim_{x \to 0^-} \frac{-x}{x} = -1 \)

RHL \( = \lim_{x \to 0^+} \frac{x}{x} = 1 \)

Since LHL \( \ne \) RHL, limit does not exist.

Answer:

LHL \( = \lim_{x \to 0^-} \frac{x}{-x} = -1 \)

RHL \( = \lim_{x \to 0^+} \frac{x}{x} = 1 \)

Since LHL \( \ne \) RHL, limit does not exist.

Answer:

At \( x=5 \), \( |x|=x \). So function behaves continuously.

\( \lim_{x \to 5} (|x| - 5) = |5| - 5 = 5 - 5 = 0 \)

Answer:

Given \( \lim_{x \to 1} f(x) = f(1) = 4 \).

LHL \( = \lim_{x \to 1^-} (a+bx) = a+b = 4 \)

RHL \( = \lim_{x \to 1^+} (b-ax) = b-a = 4 \)

Adding equations: \( (a+b) + (b-a) = 4 + 4 \Rightarrow 2b = 8 \Rightarrow b = 4 \).

Substitute \( b=4 \): \( a+4=4 \Rightarrow a=0 \).

Result: \( a=0, b=4 \)

Answer:

(i) \( \lim_{x \to a_1} f(x) = (a_1-a_1)(a_1-a_2)\dots = 0 \).

(ii) \( \lim_{x \to a} f(x) = (a-a_1)(a-a_2)\dots(a-a_n) \).

Answer:

Check at \( a=0 \):

LHL \( = \lim_{x \to 0^-} (|x|+1) = 1 \).

RHL \( = \lim_{x \to 0^+} (|x|-1) = -1 \).

Limit does not exist at \( a=0 \). For any other point \( a \ne 0 \), the function is continuous polynomial/modulus, so limit exists.

Answer: For all real \( a \ne 0 \).

Answer:

Since the limit of the denominator \( x^2-1 \) is 0 as \( x \to 1 \), for the entire limit to be finite (\( \pi \)), the limit of the numerator must also be 0.

\( \lim_{x \to 1} (f(x)-2) = 0 \Rightarrow \lim_{x \to 1} f(x) = 2 \).

Answer:

At \( x=0 \):

LHL \( = m(0)^2+n = n \).

RHL \( = n(0)+m = m \).

For limit to exist, \( m = n \).

At \( x=1 \):

LHL \( = n(1)+m = n+m \).

RHL \( = n(1)^3+m = n+m \).

Limit at \( x=1 \) exists for all integer values of \( m, n \).

Conclusion: Both limits exist for any integral values such that \( m = n \).

Answer:

Let \( f(x) = x^2 - 2 \).

Differentiating with respect to \( x \):

\( f'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(2) \)

Using the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) and derivative of a constant is 0:

\( f'(x) = 2x - 0 = 2x \)

At \( x = 10 \):

\( f'(10) = 2(10) = 20 \)

Answer:

Let \( f(x) = x \).

\( f'(x) = \frac{d}{dx}(x^1) = 1 \cdot x^{1-1} = 1 \cdot x^0 = 1 \).

At \( x = 1 \):

\( f'(1) = 1 \).

Answer:

Let \( f(x) = 99x \).

\( f'(x) = 99 \frac{d}{dx}(x) = 99(1) = 99 \).

At \( x = 100 \):

\( f'(100) = 99 \).

Answer:

Definition of First Principle: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

(i) \( f(x) = x^3 - 27 \)

\( f(x+h) = (x+h)^3 - 27 = x^3 + h^3 + 3x^2h + 3xh^2 - 27 \)

\( f'(x) = \lim_{h \to 0} \frac{(x^3 + h^3 + 3x^2h + 3xh^2 - 27) - (x^3 - 27)}{h} \)

\( = \lim_{h \to 0} \frac{h^3 + 3x^2h + 3xh^2}{h} \)

\( = \lim_{h \to 0} \frac{h(h^2 + 3x^2 + 3xh)}{h} \)

\( = \lim_{h \to 0} (h^2 + 3x^2 + 3xh) = 0 + 3x^2 + 0 = 3x^2 \)

(ii) \( f(x) = (x-1)(x-2) = x^2 - 3x + 2 \)

\( f(x+h) = (x+h)^2 - 3(x+h) + 2 = x^2 + 2xh + h^2 - 3x - 3h + 2 \)

\( f'(x) = \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 - 3x - 3h + 2) - (x^2 - 3x + 2)}{h} \)

\( = \lim_{h \to 0} \frac{2xh + h^2 - 3h}{h} \)

\( = \lim_{h \to 0} \frac{h(2x + h - 3)}{h} \)

\( = \lim_{h \to 0} (2x + h - 3) = 2x - 3 \)

(iii) \( f(x) = \frac{1}{x^2} \)

\( f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} \)

\( = \lim_{h \to 0} \frac{x^2 - (x+h)^2}{h \cdot x^2 (x+h)^2} \)

\( = \lim_{h \to 0} \frac{x^2 - (x^2 + 2xh + h^2)}{h x^2 (x+h)^2} \)

\( = \lim_{h \to 0} \frac{-2xh - h^2}{h x^2 (x+h)^2} = \lim_{h \to 0} \frac{-h(2x + h)}{h x^2 (x+h)^2} \)

\( = \lim_{h \to 0} \frac{-(2x + h)}{x^2 (x+h)^2} = \frac{-(2x + 0)}{x^2 (x)^2} = \frac{-2x}{x^4} = -\frac{2}{x^3} \)

(iv) \( f(x) = \frac{x+1}{x-1} \)

\( f'(x) = \lim_{h \to 0} \frac{\frac{(x+h)+1}{(x+h)-1} - \frac{x+1}{x-1}}{h} \)

\( = \lim_{h \to 0} \frac{1}{h} \left[ \frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{(x+h-1)(x-1)} \right] \)

Expanding the numerator:

\( [(x^2 - x + hx - h + x - 1)] - [(x^2 + hx - x + x + h - 1)] \)

\( = (x^2 + hx - 1) - (x^2 + hx + h - 1) = -h \)

Substitute back into limit:

\( = \lim_{h \to 0} \frac{-h}{h(x+h-1)(x-1)} = \lim_{h \to 0} \frac{-1}{(x+h-1)(x-1)} \)

\( = \frac{-1}{(x-1)(x-1)} = \frac{-2}{(x-1)^2} \)

Answer:

We differentiate term by term using \( \frac{d}{dx}(\frac{x^n}{n}) = \frac{1}{n} \cdot n x^{n-1} = x^{n-1} \).

\( f'(x) = \frac{d}{dx}(\frac{x^{100}}{100}) + \frac{d}{dx}(\frac{x^{99}}{99}) + \dots + \frac{d}{dx}(\frac{x^2}{2}) + \frac{d}{dx}(x) + \frac{d}{dx}(1) \)

\( f'(x) = x^{99} + x^{98} + \dots + x^1 + 1 + 0 \)

Find \( f'(1) \):

\( f'(1) = 1^{99} + 1^{98} + \dots + 1 + 1 \). Since the powers go from 1 to 99, there are 99 terms, plus the constant 1. Total 100 terms.

\( f'(1) = 100 \).

Find \( f'(0) \):

\( f'(0) = 0^{99} + 0^{98} + \dots + 0 + 1 = 1 \).

Clearly, \( 100 = 100 \times 1 \), so \( f'(1) = 100 f'(0) \). Hence Proved.

Answer:

Let \( f(x) = x^n + ax^{n-1} + a^2x^{n-2} + \dots + a^{n-1}x + a^n \).

Differentiating term by term:

• \( \frac{d}{dx}(x^n) = nx^{n-1} \)
• \( \frac{d}{dx}(ax^{n-1}) = a(n-1)x^{n-2} \)
• \( \frac{d}{dx}(a^2x^{n-2}) = a^2(n-2)x^{n-3} \)
• ...
• \( \frac{d}{dx}(a^{n-1}x) = a^{n-1}(1) = a^{n-1} \)
• \( \frac{d}{dx}(a^n) = 0 \) (Since \( a^n \) is a constant)

Combining these:

\( f'(x) = nx^{n-1} + a(n-1)x^{n-2} + a^2(n-2)x^{n-3} + \dots + a^{n-1} \).

Answer:

(i) \( f(x) = (x-a)(x-b) \)

Expand first: \( f(x) = x^2 - bx - ax + ab = x^2 - (a+b)x + ab \).

\( f'(x) = 2x - (a+b)(1) + 0 = 2x - a - b \).

(ii) \( f(x) = (ax^2+b)^2 \)

Expand: \( f(x) = a^2x^4 + 2abx^2 + b^2 \).

\( f'(x) = a^2(4x^3) + 2ab(2x) + 0 \)

\( f'(x) = 4a^2x^3 + 4abx = 4ax(ax^2+b) \).

(iii) \( f(x) = \frac{x-a}{x-b} \)

Using Quotient Rule \( [\frac{u}{v}]' = \frac{u'v - uv'}{v^2} \):

Let \( u = x-a \Rightarrow u' = 1 \).

Let \( v = x-b \Rightarrow v' = 1 \).

\( f'(x) = \frac{1(x-b) - 1(x-a)}{(x-b)^2} = \frac{x - b - x + a}{(x-b)^2} = \frac{a-b}{(x-b)^2} \).

Answer:

Using Quotient Rule:

\( u = x^n - a^n \Rightarrow u' = nx^{n-1} \).

\( v = x - a \Rightarrow v' = 1 \).

\( f'(x) = \frac{(nx^{n-1})(x-a) - (x^n - a^n)(1)}{(x-a)^2} \)

\( = \frac{nx^n - anx^{n-1} - x^n + a^n}{(x-a)^2} \)

\( = \frac{(n-1)x^n - anx^{n-1} + a^n}{(x-a)^2} \).

Answer:

(i) \( \frac{d}{dx}(2x - \frac{3}{4}) = 2(1) - 0 = 2 \).

(ii) Using Product Rule \( (uv)' = u'v + uv' \):

\( u = 5x^3+3x-1 \Rightarrow u' = 15x^2+3 \)

\( v = x-1 \Rightarrow v' = 1 \)

\( f'(x) = (15x^2+3)(x-1) + (5x^3+3x-1)(1) \)

\( = 15x^3 - 15x^2 + 3x - 3 + 5x^3 + 3x - 1 \)

\( = 20x^3 - 15x^2 + 6x - 4 \).

(iii) Expand: \( f(x) = 5x^{-3} + 3x^{-2} \).

\( f'(x) = 5(-3x^{-4}) + 3(-2x^{-3}) = -15x^{-4} - 6x^{-3} \).

(iv) Expand: \( f(x) = 3x^5 - 6x^{5-9} = 3x^5 - 6x^{-4} \).

\( f'(x) = 3(5x^4) - 6(-4x^{-5}) = 15x^4 + 24x^{-5} \).

(v) Expand: \( f(x) = 3x^{-4} - 4x^{-9} \).

\( f'(x) = 3(-4x^{-5}) - 4(-9x^{-10}) = -12x^{-5} + 36x^{-10} \).

(vi) Differentiate separately using Quotient Rule:

\( \frac{d}{dx}(\frac{2}{x+1}) = \frac{0(x+1) - 2(1)}{(x+1)^2} = \frac{-2}{(x+1)^2} \).

\( \frac{d}{dx}(\frac{x^2}{3x-1}) = \frac{2x(3x-1) - x^2(3)}{(3x-1)^2} = \frac{6x^2-2x-3x^2}{(3x-1)^2} = \frac{3x^2-2x}{(3x-1)^2} \).

Result: \( \frac{-2}{(x+1)^2} - \frac{3x^2-2x}{(3x-1)^2} \).

Answer:

\( f(x) = \cos x \).

\( f'(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} \)

Using the identity \( \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} \):

\( \cos(x+h) - \cos x = -2 \sin\left(\frac{2x+h}{2}\right) \sin\left(\frac{h}{2}\right) \)

\( f'(x) = \lim_{h \to 0} \frac{-2 \sin(x + h/2) \sin(h/2)}{h} \)

\( = -\lim_{h \to 0} \sin(x + h/2) \cdot \lim_{h \to 0} \frac{\sin(h/2)}{h/2} \)

Since \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \):

\( = - \sin(x+0) \cdot 1 = -\sin x \).

Answer:

(i) \( f(x) = \sin x \cos x \)

Using Product Rule: \( (\cos x)(\cos x) + (\sin x)(-\sin x) = \cos^2 x - \sin^2 x = \cos 2x \).

(ii) \( f(x) = \sec x \)

Since \( \sec x = \frac{1}{\cos x} \), using quotient rule or standard formula:

\( f'(x) = \sec x \tan x \).

(iii) \( f(x) = 5\sec x + 4\cos x \)

\( f'(x) = 5(\sec x \tan x) + 4(-\sin x) = 5\sec x \tan x - 4\sin x \).

(iv) \( f(x) = \csc x \)

Standard formula: \( f'(x) = -\csc x \cot x \).

(v) \( f(x) = 3\cot x + 5\csc x \)

\( f'(x) = 3(-\csc^2 x) + 5(-\csc x \cot x) = -3\csc^2 x - 5\csc x \cot x \).

(vi) \( f(x) = 5\sin x - 6\cos x + 7 \)

\( f'(x) = 5(\cos x) - 6(-\sin x) + 0 = 5\cos x + 6\sin x \).

(vii) \( f(x) = 2\tan x - 7\sec x \)

\( f'(x) = 2(\sec^2 x) - 7(\sec x \tan x) = 2\sec^2 x - 7\sec x \tan x \).

Answer:

(i) \( f(x) = -x \)

\( f'(x) = \lim_{h \to 0} \frac{-(x+h) - (-x)}{h} = \lim_{h \to 0} \frac{-x-h+x}{h} = \lim_{h \to 0} \frac{-h}{h} = -1 \).

(ii) \( f(x) = (-x)^{-1} = -\frac{1}{x} \)

\( f'(x) = \lim_{h \to 0} \frac{-\frac{1}{x+h} - (-\frac{1}{x})}{h} = \lim_{h \to 0} \frac{\frac{1}{x} - \frac{1}{x+h}}{h} \)

\( = \lim_{h \to 0} \frac{(x+h) - x}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{h}{h x(x+h)} = \lim_{h \to 0} \frac{1}{x(x+h)} = \frac{1}{x^2} \).

(iii) \( f(x) = \sin(x+1) \)

\( f'(x) = \lim_{h \to 0} \frac{\sin(x+h+1) - \sin(x+1)}{h} \)

Use \( \sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \):

\( A = x+h+1, B = x+1 \Rightarrow \frac{A+B}{2} = x+1+\frac{h}{2}, \frac{A-B}{2} = \frac{h}{2} \).

\( = \lim_{h \to 0} \frac{2\cos(x+1+h/2)\sin(h/2)}{h} \)

\( = \lim_{h \to 0} \cos(x+1+h/2) \cdot \frac{\sin(h/2)}{h/2} = \cos(x+1) \cdot 1 = \cos(x+1) \).

(iv) \( f(x) = \cos(x - \pi/8) \)

Similar to the cosine derivative derivation:

\( f'(x) = -\sin(x - \pi/8) \).

Answer:

\( f(x) = x + a \).

\( f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(a) = 1 + 0 = 1 \).

Answer:

Expand first: \( f(x) = pr + psx + \frac{qr}{x} + qs \).

\( f'(x) = 0 + ps(1) + qr(-x^{-2}) + 0 \)

\( f'(x) = ps - \frac{qr}{x^2} \).

Answer:

Let \( u = ax+b \) and \( v = (cx+d)^2 \).

\( u' = a \).

\( v' = 2(cx+d) \cdot c = 2c(cx+d) \) (Using Chain Rule or expansion).

Using Product Rule \( u'v + uv' \):

\( f'(x) = a(cx+d)^2 + (ax+b) \cdot 2c(cx+d) \)

Factor out \( (cx+d) \):

\( = (cx+d) [ a(cx+d) + 2c(ax+b) ] \)

\( = (cx+d) [ acx + ad + 2acx + 2bc ] \)

\( = (cx+d) (3acx + ad + 2bc) \).

Answer:

Using Quotient Rule:

\( u = ax+b \Rightarrow u' = a \)

\( v = cx+d \Rightarrow v' = c \)

\( f'(x) = \frac{a(cx+d) - c(ax+b)}{(cx+d)^2} \)

\( = \frac{acx + ad - acx - bc}{(cx+d)^2} = \frac{ad-bc}{(cx+d)^2} \).

Answer:

Simplify first: \( f(x) = \frac{(x+1)/x}{(x-1)/x} = \frac{x+1}{x-1} \).

Using Quotient Rule:

\( f'(x) = \frac{1(x-1) - 1(x+1)}{(x-1)^2} = \frac{x-1-x-1}{(x-1)^2} = \frac{-2}{(x-1)^2} \).

Answer:

Using Quotient Rule with \( u=1, u'=0 \):

\( f'(x) = \frac{0(ax^2+bx+c) - (2ax+b)(1)}{(ax^2+bx+c)^2} \)

\( = \frac{-(2ax+b)}{(ax^2+bx+c)^2} \).

Answer:

Using Quotient Rule:

Numerator deriv: \( a \). Denominator deriv: \( 2px+q \).

\( f'(x) = \frac{a(px^2+qx+r) - (2px+q)(ax+b)}{(px^2+qx+r)^2} \)

\( = \frac{apx^2+aqx+ar - (2apx^2 + 2bpx + aqx + bq)}{(px^2+qx+r)^2} \)

\( = \frac{-apx^2 - 2bpx + ar - bq}{(px^2+qx+r)^2} \).

Answer:

Using Quotient Rule:

\( f'(x) = \frac{(2px+q)(ax+b) - a(px^2+qx+r)}{(ax+b)^2} \)

\( = \frac{2apx^2 + 2bpx + aqx + bq - apx^2 - aqx - ar}{(ax+b)^2} \)

\( = \frac{apx^2 + 2bpx + bq - ar}{(ax+b)^2} \).

Answer:

Rewrite: \( f(x) = ax^{-4} - bx^{-2} + \cos x \).

\( f'(x) = a(-4x^{-5}) - b(-2x^{-3}) - \sin x \)

\( = -\frac{4a}{x^5} + \frac{2b}{x^3} - \sin x \).

Answer:

\( f(x) = 4x^{1/2} - 2 \).

\( f'(x) = 4(\frac{1}{2}x^{-1/2}) - 0 = 2x^{-1/2} = \frac{2}{\sqrt{x}} \).

Answer:

Using Chain Rule: \( \frac{d}{dx}(u^n) = nu^{n-1} \cdot u' \).

\( f'(x) = n(ax+b)^{n-1} \cdot \frac{d}{dx}(ax+b) \)

\( = n(ax+b)^{n-1} \cdot a = an(ax+b)^{n-1} \).

Answer:

Using Product Rule and Chain Rule:

\( u = (ax+b)^n \Rightarrow u' = na(ax+b)^{n-1} \).

\( v = (cx+d)^m \Rightarrow v' = mc(cx+d)^{m-1} \).

\( f'(x) = na(ax+b)^{n-1}(cx+d)^m + mc(cx+d)^{m-1}(ax+b)^n \).

Factoring \( (ax+b)^{n-1}(cx+d)^{m-1} \):

\( = (ax+b)^{n-1}(cx+d)^{m-1} [ na(cx+d) + mc(ax+b) ] \).

Answer:

Using Chain Rule or sum formula.

\( f(x) = \sin x \cos a + \cos x \sin a \).

\( f'(x) = \cos x \cos a - \sin x \sin a = \cos(x+a) \).

Answer:

Product Rule: \( u = \csc x, v = \cot x \).

\( u' = -\csc x \cot x, v' = -\csc^2 x \).

\( f'(x) = (-\csc x \cot x)(\cot x) + (\csc x)(-\csc^2 x) \)

\( = -\csc x \cot^2 x - \csc^3 x = -\csc x (\cot^2 x + \csc^2 x) \).

Answer:

Using Quotient Rule:

\( f'(x) = \frac{-\sin x(1+\sin x) - \cos x(\cos x)}{(1+\sin x)^2} \)

\( = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2} = \frac{-\sin x - (\sin^2 x + \cos^2 x)}{(1+\sin x)^2} \)

\( = \frac{-\sin x - 1}{(1+\sin x)^2} = \frac{-(1+\sin x)}{(1+\sin x)^2} = \frac{-1}{1+\sin x} \).

Answer:

Using Quotient Rule:

\( u = \sin x + \cos x \Rightarrow u' = \cos x - \sin x \).

\( v = \sin x - \cos x \Rightarrow v' = \cos x + \sin x \).

\( f'(x) = \frac{(\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2} \)

\( = \frac{-(\sin x - \cos x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2} \)

\( = \frac{-( \sin^2 x + \cos^2 x - 2\sin x \cos x + \sin^2 x + \cos^2 x + 2\sin x \cos x )}{(\sin x - \cos x)^2} \)

\( = \frac{-(1 + 1)}{(\sin x - \cos x)^2} = \frac{-2}{(\sin x - \cos x)^2} \).

Answer:

Simplify first: \( f(x) = \frac{1/\cos x - 1}{1/\cos x + 1} = \frac{1 - \cos x}{1 + \cos x} \).

Using half-angle formulas: \( f(x) = \frac{2\sin^2(x/2)}{2\cos^2(x/2)} = \tan^2(x/2) \).

\( f'(x) = 2\tan(x/2) \cdot \sec^2(x/2) \cdot \frac{1}{2} = \tan(x/2) \sec^2(x/2) \).

Answer:

Using Chain Rule for \( (\sin x)^n \):

\( f'(x) = n(\sin x)^{n-1} \cdot \frac{d}{dx}(\sin x) = n \sin^{n-1} x \cos x \).

Answer:

Using Quotient Rule:

\( u = a+b\sin x \Rightarrow u' = b\cos x \).

\( v = c+d\cos x \Rightarrow v' = -d\sin x \).

\( f'(x) = \frac{b\cos x(c+d\cos x) - (a+b\sin x)(-d\sin x)}{(c+d\cos x)^2} \)

\( = \frac{bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x}{(c+d\cos x)^2} \)

\( = \frac{bc\cos x + ad\sin x + bd(\cos^2 x + \sin^2 x)}{(c+d\cos x)^2} \)

\( = \frac{bc\cos x + ad\sin x + bd}{(c+d\cos x)^2} \).

Answer:

Using Quotient Rule:

\( f'(x) = \frac{\cos(x+a)\cos x - \sin(x+a)(-\sin x)}{\cos^2 x} \)

\( = \frac{\cos(x+a)\cos x + \sin(x+a)\sin x}{\cos^2 x} \)

Using \( \cos(A-B) \):

\( = \frac{\cos(x+a-x)}{\cos^2 x} = \frac{\cos a}{\cos^2 x} \).

Answer:

Using Product Rule:

\( f'(x) = 4x^3(5\sin x - 3\cos x) + x^4(5\cos x - 3(-\sin x)) \)

\( = x^3 [ 20\sin x - 12\cos x + 5x\cos x + 3x\sin x ] \).

Answer:

Using Product Rule:

\( f'(x) = 2x \cos x + (x^2+1)(-\sin x) = 2x\cos x - (x^2+1)\sin x \).

Answer:

Using Product Rule:

\( u = ax^2+\sin x \Rightarrow u' = 2ax+\cos x \).

\( v = p+q\cos x \Rightarrow v' = -q\sin x \).

\( f'(x) = (2ax+\cos x)(p+q\cos x) + (ax^2+\sin x)(-q\sin x) \).

Answer:

Using Product Rule:

\( f'(x) = (1-\sin x)(x-\tan x) + (x+\cos x)(1-\sec^2 x) \).

Answer:

Using Quotient Rule:

\( f'(x) = \frac{(4+5\cos x)(3x+7\cos x) - (3-7\sin x)(4x+5\sin x)}{(3x+7\cos x)^2} \).

Answer:

\( \cos(\pi/4) = \frac{1}{\sqrt{2}} \) is a constant.

\( f(x) = \frac{1}{\sqrt{2}} \frac{x^2}{\sin x} \).

\( f'(x) = \frac{1}{\sqrt{2}} \left[ \frac{2x\sin x - x^2\cos x}{\sin^2 x} \right] \).

Answer:

Using Quotient Rule:

\( f'(x) = \frac{1(1+\tan x) - x(\sec^2 x)}{(1+\tan x)^2} \)

\( = \frac{1+\tan x - x\sec^2 x}{(1+\tan x)^2} \).

Answer:

Using Product Rule:

\( f'(x) = (1+\sec x \tan x)(x-\tan x) + (x+\sec x)(1-\sec^2 x) \).

Answer:

\( f'(x) = \frac{1(\sin^n x) - x(n\sin^{n-1}x \cos x)}{(\sin^n x)^2} \)

Divide numerator and denominator by \( \sin^{n-1}x \):

\( = \frac{\sin x - nx\cos x}{\sin^{n+1} x} \).