Chapter 12: Limits and Derivatives

Ans: Let $f\left( x \right) = \sin x\cos x$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {x + h} \right)\cos \left( {x + h} \right) - \sin x\cos x}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\sin \left( {x + h} \right)\cos \left( {x + h} \right) - 2\sin x\cos x} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {\sin 2\left( {x + h} \right) - \sin 2x} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\cos \frac{{2x + 2h + 2x}}{2}.\sin \frac{{2x + 2h - 2x}}{2}} \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {2\cos \frac{{4x + 2h}}{2}.\sin \frac{{2h}}{2}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\left[ {\cos \left( {2x + h} \right)\sinh } \right]$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \cos \left( {2x + h} \right).\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}$

$\Rightarrow \cos \left( {2x + h} \right) \times 1 = \cos 2x$

Ans: Let $f\left( x \right) = \sec x$

Accordingly, from the first principle

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\sec \left( {x + h} \right) - \sec x}}{h}$

$\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos \left( {x + h} \right)}} - \frac{1}{{\cos x}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right]$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right]$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{{2h}}\frac{{\left[ { - 2\sin \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right]}}{{\cos \left( {x + h} \right)}}$

$\Rightarrow \frac{1}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}$

$\Rightarrow \frac{1}{{\cos x}} \times 1 \times \frac{{\sin x}}{{\cos x}} = \sec x\tan x$

Ans: Let, 

$f\left( x \right) = 5\sec x + 4\cos x$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \frac{{\lim }}{{h \to 0}}\frac{{5\sec \left( {x + h} \right) + 4\cos \left( {x + h} \right) - \left[ {5\sec x + 4\cos x} \right]}}{h}$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sec \left( {x + h} \right) - \sec x} \right]}}{h} + 4\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\cos \left( {x + h} \right) - \cos x} \right]}}{h}$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos \left( {x + h} \right)}} - \frac{1}{{\cos x}}} \right] + 4\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \left( {x + h} \right) - \cos x} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right] + 4\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos x\cosh  - \sin x\sinh  - \cos x} \right]$

$ \Rightarrow \frac{5}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right] + 4\left[ { - \cos x\mathop {\lim }\limits_{h \to 0} \frac{{\left( {1 - \cos x} \right)}}{h} - \sin x\mathop {\lim \frac{{\sinh }}{h}}\limits_{h \to 0} } \right]$

$ \Rightarrow \frac{5}{{\cos x}}\mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sin \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right]}}{{\cos \left( {x + h} \right)}} + 4\left[ { - \cos x\left( 0 \right) - \sin x\left( 1 \right)} \right]$

$ \Rightarrow \frac{5}{{\cos x}}\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos \left( {x + h} \right)}}} \right] - 4\sin x$

$ \Rightarrow \frac{5}{{\cos x}} \times \frac{{\sin x}}{{\cos x}} \times 1 - 4\sin x = 5\sec x\tan x - 4\sin x$

Ans: Let $f\left( x \right) = \cos ecx$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos ec\left( {x + h} \right) - \cos ecx} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\sin \left( {x + h} \right)}} - \frac{1}{{\sin x}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x - \sin \left( {x + h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{\left[ { - \cos \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right]}}{{\sin x\sin \left( {x + h} \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - \cos \left( {\frac{{2x + h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right)\mathop {\lim }\limits_{\frac{h}{2} \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}$

$ \Rightarrow \left( {\frac{{ - \cos x}}{{\sin x\sin x}}} \right) \times 1 =  - \cos ecx\cot x$

Ans: Let $f\left( x \right) = 3\cot x + 5\cos ecx$

Accordingly, from the first principle

 $f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {3\cot \left( {x + h} \right) + 5\cos ec\left( {x + h} \right) - 3\cot x - 5\cos ecx} \right]$

\[ \Rightarrow 3\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cot \left( {x + h} \right) - \cot x} \right] + 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos ec\left( {x + h} \right) - \cos ecx} \right]\,\,\,\,\,\,\,\,\,\,....\left( 1 \right)\]

Now, $\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cot \left( {x + h} \right) - \cot x} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos \left( {x + h} \right)}}{{\sin \left( {x + h} \right)}} - \frac{{\cos x}}{{\sin x}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos \left( {x + h} \right)\sin x - \cos x\sin \left( {x + h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin \left( {x - x - h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin \left( { - h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}.\mathop {\lim }\limits_{h \to 0} \left[ {\frac{1}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow  - 1 \times \frac{1}{{\sin x\sin \left( {x + h} \right)}} = \frac{{ - 1}}{{{{\sin }^2}x}} =  - \cos e{c^2}x\,\,\,\,\,\,......(2)$

$\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos ec\left( {x + h} \right) - \cos ecx} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\sin \left( {x + h} \right)}} - \frac{1}{{\sin x}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x - \sin \left( {x + h} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right]$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{2\cos \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right] = \mathop {\lim }\limits_{h \to 0} \frac{{ - \cos \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{{ - h}}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}}}{{\sin x\sin \left( {x + h} \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - \cos \left( {\frac{{2x + h}}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}} \right)\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}} = \left( {\frac{{ - \cos x}}{{\sin x\sin x}}} \right).1$

$ \Rightarrow  - \cos ecx\cot x\,\,\,\,\,\,......(3)$

From (1), (2), and (3), we obtain

$f'\left( x \right) =  - 3\cos e{c^2}x - 5\cos ecx\cot x$

Ans: Let $f\left( x \right) = 5\sin x - 6\cos x + 7$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {5\sin \left( {x + h} \right) - 6\cos \left( {x + h} \right) + 7 - 5\sin x + 6\cos x - 7} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\sin \left( {x + h} \right) - \sin x} \right] - 6\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \left( {x + h} \right) - \cos x} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\cos \left( {\frac{{x + h + x}}{2}} \right).\sin \left( {\frac{{x + h - x}}{2}} \right)} \right] - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{\cos x\cosh  - \sin x\sinh  - \cos x}}{h}} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\cos \left( {\frac{{2x + h}}{2}} \right).\sin \left( {\frac{h}{2}} \right)} \right] - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cosh } \right) - \sin x\sinh }}{h}} \right]$

$ \Rightarrow 5\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\cos \left( {\frac{{2x + h}}{2}} \right)\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right] - 6\mathop {\lim }\limits_{h \to 0} \left[ {\frac{{ - \cos x\left( {1 - \cosh } \right)}}{h} - \frac{{\sin x\sinh }}{h}} \right]$

$ \Rightarrow 5\left[ {\mathop {\lim }\limits_{h \to 0} \cos \left( {\frac{{2x + h}}{2}} \right)} \right]\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right] - 6\left[ { - \cos x\left( {\mathop {\lim }\limits_{h \to 0} \frac{{1 - \cosh }}{h}} \right) - \sin x\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}} \right)} \right]$

$ \Rightarrow 5\cos x.1 - 6\left[ {\left( { - \cos x} \right).\left( 0 \right) - \sin x.1} \right]$

$ \Rightarrow 5\cos x + 6\sin x$

Ans: Let $f\left( x \right) = 2\tan x - 7\sec x$

Accordingly, from the first principle,

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$

$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\tan \left( {x + h} \right) - 7\sec \left( {x + h} \right) - 2\tan x + 7\sec x}\right]$

  $ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\tan \left( {x + h} \right) - \tan x} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\sec \left( {x + h} \right) - \sec x} \right]$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin \left( {x + h} \right)}}{{\cos \left( {x + h} \right)}} - \frac{{\sin x}}{{\cos x}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{1}{{\cos ec\left( {x + h} \right)}} - \frac{1}{{\cos ecx}}} \right]$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x\sin \left( {x + h} \right) - \sin x\cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\cos x - \cos \left( {x + h} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$ \Rightarrow 2\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{\sin x + h - x}}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{x + x + h}}{2}} \right)\sin \left( {\frac{{x - x - h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$ \Rightarrow 2\left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{\sinh }}{h}} \right)\frac{1}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {\frac{{ - 2\sin \left( {\frac{{2x + h}}{2}} \right)\sin \left( {\frac{{ - h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right]$

$ \Rightarrow 2\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h}} \right)\left[ {\mathop {\lim }\limits_{h \to 0} \frac{1}{{\cos x\cos \left( {x + h} \right)}}} \right] - 7\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}} \right)\left( {\mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {\frac{{2x + h}}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}} \right)$

$ \Rightarrow 2 \times 1 \times 1 \times \frac{1}{{\cos x\cos x}} - 7\left( {1 \times \frac{{\sin x}}{{\cos x\cos x}}} \right) = 2{\sec ^2}x - 7\sec x\tan x$