ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 4- Worked Out Examples

Saturday, 2 August 2014

CHAPTER 4- Worked Out Examples

Evaluate the derivatives of the following functions by first principles:

	$(a)f\left( x \right) = \log \left( {\sin x} \right)$
	$(b) f\left( x \right) = \sin \left( {\log x} \right)$
	$(c) f\left( x \right) = \sqrt[3]{{\sin x}}$
	$(d)f\left( x \right) = {e^{{x^2}}}$

Solution: 1-(a)

These functions are all compositions of multiple functions and can easily be differentiated using the Chain rule. However, our purpose here is to differentiate them using first principles. This exercise, instead of being considered futile, should be seen as an application of your algebraic manipulation skills.

	$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left( {\sin \left( {x + h} \right)} \right) - \log \left( {\sin x} \right)}}{h}$
	$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left( {\dfrac{{\sin \left( {x + h} \right)}}{{\sin x}}} \right)}}{h}$
	$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left( {\dfrac{{\sin x\cos h + \cos x\sin h}}{{\sin x}}} \right)}}{h}$
	$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left( {\cos h + \cot x\sin h} \right)}}{h}$
	$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left\{ {\cos h\left( {1 + \cot x\tan h} \right)} \right\}}}{h}$
	$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left( {\cos h} \right)}}{h} + \mathop {\lim }\limits_{h \to 0} \dfrac{{\log \left( {1 + \cot x\tan h} \right)}}{h}$
	$\mathop {\lim }\limits_{h \to 0} \dfrac{{\log \{ 1 + (\cosh - 1)}}{h} + \mathop {\lim }\limits_{h \to 0}\overbrace{ \dfrac{{\log (1 + \cot x\tanh )}}{{\cot x\tanh }}} ^{Comment}.\dfrac{{\cot x\tanh }}{h}$ Comment: This limit is $1$
	$\mathop {\lim }\limits_{h \to 0}\overbrace{ \dfrac{{\log \{ 1 + (\cosh - 1)\} }}{{\cosh - 1}}}^{Comment}\dfrac{{\cosh - 1}}{h} + \cot x$ Comment: This limit is $1$
	$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos h - 1}}{h} + \cot x$
	$= 0 + \cot x$
	$= \cot x$

Solution: 1-(b)

	$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {\log \left( {x + h} \right)} \right) - \sin \left( {\log x} \right)}}{h}$
	$= 2\mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left\{ {\dfrac{{\log \left( {x + h} \right) + \log \left( x \right)}}{2}} \right\}\sin \left\{ {\dfrac{{\log \left( {x + h} \right) - \log \left( x \right)}}{2}} \right\}}}{h}$
	$= 2\cos \left( {\log \left( x \right)} \right)\mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{\sin \left\{ {\dfrac{1}{2}\log \left( {1 + \dfrac{h}{x}} \right)} \right\}}}{h}} \right\}$
	$= 2\cos \left( {\log \left( x \right)} \right)\mathop {\lim }\limits_{h \to 0} \left[ {\left\{ {\dfrac{{\sin \left\{ {\dfrac{1}{2}\log \left( {1 + \dfrac{h}{x}} \right)} \right\}}}{{\dfrac{1}{2}\log \left( {1 + \dfrac{h}{x}} \right)}}} \right\}\, \cdot \,\dfrac{{\dfrac{1}{2}\log \left( {1 + \dfrac{h}{x}} \right)}}{h}} \right]$
	$= 2\cos (\log (x))\mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{1}{2}\dfrac{{\log (1 + \dfrac{h}{x}}}{{x.h/x}}} \right\}$
	$= \dfrac{{\cos \left( {\log \left( x \right)} \right)}}{x}$

Solution: 1-(c)

	$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt[3]{{\sin \left( {x + h} \right)}} - \sqrt[3]{{\sin x}}}}{h}$
	$= \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {x + h} \right) - \sin \left( x \right)}}{h} \cdot \dfrac{1}{{{{\left[ {\sin \left( {x + h} \right)} \right]}^{2/3}} + {{\left( {\sin x} \right)}^{2/3}} + {{\left[ {\sin x \cdot \sin \left( {x + h} \right)} \right]}^{1/3}}}}$
	{This step was accomplished by rationalization}
	$= \dfrac{{\cos x}}{{3 \cdot {{\left( {\sin x} \right)}^{2/3}}}}$

Solution: 1-(d)

	$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{{{(x + h)}^2}}} - {e^{{x^2}}}}}{h}$
	$= \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{{x^2} + {h^2} + 2xh}} - {e^{{x^2}}}}}{h}$
	$= {e^{{x^2}}}\mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{{h^2} + 2xh}} - 1}}{h}$
	${\rm{ = }}{{\rm{e}}^{{x^2}}}\mathop {\lim }\limits_{h \to 0} \left\{ {\overbrace{\dfrac{{{e^{{h^2} + 2xh}} - 1}}{{{h^2} + 2xh}}} ^{Comment}.\dfrac{{{h^2} + 2xh}}{h}} \right\}{\rm{e 1}}$ Comment: The limiting value of this ratio will be $1$
	$= {e^{{x^2}}} \cdot 2x$

Example: 2

	(a) $f\left( x \right) = \left( {{x^2} + 3x} \right){\sin ^{ - 1}}x$
	(b) $f\left( x \right) = \tan \left( {\dfrac{{x + 1}}{{x + 2}}} \right)$
	(c) $f\left( x \right) = {\cos ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right) + \dfrac{{{x^2} + 2x + 3}}{{x + 2}}$
	(d) $f\left( x \right) = \sqrt {4 - {{\tan }^{ - 1}}{x^2}}$

Solution: 2-(a)

This function can be differentiated by the Product Rule (Rule – $3$ )

	$\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \left( {{x^2} + 3x} \right)\dfrac{{d\left( {{{\sin }^{ - 1}}x} \right)}}{{dx}} + {\sin ^{ - 1}}x\dfrac{{d\left( {{x^2} + 3x} \right)}}{{dx}}$
	$= \dfrac{{\left( {{x^2} + 3x} \right)}}{{\sqrt {1 - {x^2}} }} + \left( {{{\sin }^{ - 1}}x} \right) \cdot \left( {2x + 3} \right)$

Solution: 2-(b)

This function will be differentiated by the Chain rule (Rule $5$ ); the argument $\left( {\dfrac{{x + 1}}{{x + 2}}} \right)$ can be differentiated by the Quotient rule (Rule $4$ ):

	$\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{{d\left\{ {\tan \left( {\dfrac{{x + 1}}{{x + 2}}} \right)} \right\}}}{{d\left( {\dfrac{{x + 1}}{{x + 2}}} \right)}} \cdot \dfrac{{d\left( {\dfrac{{x + 1}}{{x + 2}}} \right)}}{{dx}}$
	$= {\sec ^2}\left( {\dfrac{{x + 1}}{{x + 2}}} \right) \cdot \dfrac{{\left( {x + 2} \right) \cdot \left( 1 \right) - \left( {x + 1} \right) \cdot \left( 1 \right)}}{{{{\left( {x + 2} \right)}^2}}}$
	$=\dfrac{1}{{{{\left( {x + 2} \right)}^2}}} \cdot {\sec ^2}\left( {\dfrac{{x + 1}}{{x + 2}}} \right)$

Solution: 2-(c)

Here, we will have to use a combination of these techniques as follows:

	$\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \left\{ {\dfrac{{ - 1}}{{1 - {{\left( {\sqrt {1 - {x^2}} } \right)}^2}}}} \right\} \cdot \dfrac{{ - x}}{{\sqrt {1 - {x^2}} }} + \dfrac{{\left( {x + 2} \right)\left( {2x + 2} \right) - \left( {{x^2} + 2x + 3} \right) \cdot \left( 1 \right)}}{{{{\left( {x + 2} \right)}^2}}}$
	$= \dfrac{1}{{x\sqrt {1 - {x^2}} }} + \dfrac{{{x^2} + 4x + 1}}{{{{\left( {x + 2} \right)}^2}}}$

You are urged to work out the solution on your own.

Solution: 2-(d)

$\dfrac{{d\left( {f\left( x \right)} \right)}}{{dx}} = \dfrac{1}{{2\sqrt {4 - {{\tan }^{ - 1}}{x^2}} }} \cdot \dfrac{{ - 1}}{{1 + {x^4}}} \cdot 2x$

$= \dfrac{{ - x}}{{\left( {1 + {x^4}} \right)\sqrt {4 - {{\tan }^{ - 1}}{x^2}} }}$

Example: 3

If $f'(x) = g(x)$ , find the derivative of ${f^{ - 1}}(x)$

Solution: 3

To evaluate the required derivative, we can apply the chain rule on the relation:

	$f\left( {{f^{ - 1}}\left( x \right)} \right) = x$
	$\Rightarrow \dfrac{d}{{dx}}\left\{ {f\left( {{f^{ - 1}}\left( x \right)} \right)} \right\} = \dfrac{d}{{dx}}\left( x \right) = 1$
	$\Rightarrow f'\left( {{f^{ - 1}}\left( x \right)} \right)\dfrac{d}{{dx}}\left( {{f^{ - 1}}\left( x \right)} \right) = 1$
	$\Rightarrow \dfrac{d}{{dx}}\left( {{f^{ - 1}}\left( x \right)} \right)\dfrac{1}{{f'\left( {{f^{ - 1}}\left( x \right)} \right)}} = \dfrac{1}{{g\left( {{f^{ - 1}}\left( x \right)} \right)}}$

For example, we know that

	$\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
	$\Rightarrow \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\cos \left( {{{\sin }^{ - 1}}\left( x \right)} \right)}}$
	$= \dfrac{1}{{\cos \left( {{{\cos }^{ - 1}}\sqrt {1 - {x^2}} } \right)}}$
	$= \dfrac{1}{{\sqrt {1 - {x^2}} }}$

In this way, we can evaluate the derivative of any inverse function, given the derivative of the original function.

Saturday, 2 August 2014

CHAPTER 4- Worked Out Examples

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