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Saturday, 2 August 2014

CHAPTER 2- Differentiation of Standard Functions

By now, the meaning and geometrical significance of differentiation should be pretty clear to you. We will use this knowledge to evaluate the derivatives of some standard functions in this section.
You will notice that while differentiating these functions, we will only use the expression for the RHD; we could equivalently use the LHD also since all the functions we will be concerned with in this section are differentiable (except at discontinuous points); that the LHD and RHD are equal for each of these functions at a given x can be easily verified.
(1)f(x) = k
 f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{k - k}}{h}
= 0
This is intuitively true also since the graph for a constant function is a horizontal line.
(2)f(x) = x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {x + h} \right) - x}}{h}
=1
This corresponds to the fact that the line f(x) = x is inclined at {45^ \circ } to the x –axis (and \tan {45^ \circ } is 1).
(3)f(x) = mx + c
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
=\mathop{\lim }\limits_{h \to 0} \dfrac{{m\left( {x + h} \right) + c-\left( {mx + c} \right)}}{h}
=m
This is again a straight forward result: ‘m’ is the slope of f\left( x \right) = mx + c so it must equal f'\left( x \right).
(4)f(x) = {x^2}
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {x + h} \right)}^2} - {x^2}}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{{h^2} + 2xh}}{h}
= 2x
We have already obtained this result earlier.
(5)f(x) = {x^n}
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {x + h} \right)}^n} - {x^n}}}{h}
 = \mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{{x^n}{{\left( {1 + \dfrac{h}{x}} \right)}^n} - {x^n}}}{h}} \right\}
 = {x^n}\mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{{{\left( {1 + \dfrac{h}{x}} \right)}^n} - 1}}{h}} \right\}
 = {x^n}\mathop {\lim }\limits_{h \to 0} \dfrac{{\left\{ {1 + \dfrac{{nh}}{x} + n\dfrac{{\left( {n - 1} \right)}}{{2!}}\dfrac{{{h^2}}}{{{x^2}}} + \ldots} \right\} - 1}}{h}
[By the binomial expansion]
 = {x^n} \cdot \dfrac{n}{x} = n{x^{n - 1}}
So, for example,
\dfrac{{d({x^2})}}{{dx}} = 2.{x^{2 - 1}} = 2x and \dfrac{{d({x^3})}}{{dx}} = 3.{x^{3 - 1}} = 3{x^2} and so on
(6)f(x) = \sin x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin \left( {x + h} \right) - \sin x}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{2\cos \left( {x + \dfrac{h}{2}} \right)\sin \left( {\dfrac{h}{2}} \right)}}{h}
 = \mathop {\lim }\limits_{h \to 0} \left\{ {\cos \left( {x + \dfrac{h}{2}} \right) \cdot \dfrac{{\sin \left( {\dfrac{h}{2}} \right)}}{{\left( {\dfrac{h}{2}} \right)}}} \right\}
 = \cos x
(7)f(x) = \cos x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos \left( {x + h} \right) - \cos x}}{h}
 = \mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{ - 2\sin \left( {x + \dfrac{h}{2}} \right)\sin \dfrac{h}{2}}}{h}} \right\}
 = \mathop {\lim }\limits_{h \to 0} \left\{ { - \sin \left( {x + \dfrac{h}{2}} \right) \cdot \dfrac{{\sin \left( {\dfrac{h}{2}} \right)}}{{\left( {\dfrac{h}{2}} \right)}}} \right\}
 =  - \sin x
(8)f(x) = \tan x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{\tan \left( {x + h} \right) - \tan x}}{h}
 = \mathop {\lim }\limits_{h \to 0} \left[ {\dfrac{{\tan \,h}}{h} \cdot \left\{ {1 + \tan x \cdot \tan \left( {x + h} \right)} \right\}} \right]
1 + {\tan ^2}x
 = {\sec ^2}x
(9)f(x) = \sec x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{\sec \left( {x + h} \right) - \sec x}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x - \cos \left( {x + h} \right)}}{{h\cos x\cos \left( {x + h} \right)}}
 = \mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{2\sin \left( {x + \dfrac{h}{2}} \right)\sin \dfrac{h}{2}}}{{h\cos x\cos \left( {x + h} \right)}}} \right\}
 = \mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{\sin \left( {x + \dfrac{h}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}\left( {\dfrac{{\sin \dfrac{h}{2}}}{{\dfrac{h}{2}}}} \right)} \right\}
 = \sec x\tan x
(10)f(x) = \cos ec x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{{\mathop{\rm cosec}\nolimits} \left( {x + h} \right) - {\mathop{\rm cosec}\nolimits} x}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin x - \sin \left( {x + h} \right)}}{{h\sin x\sin \left( {x + h} \right)}}
 = \mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{ - 2\cos \left( {x + \dfrac{h}{2}} \right)\sin \dfrac{h}{2}}}{{h\sin x\sin \left( {x + h} \right)}}} \right\}
 = \mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{ - \cos \left( {x + \dfrac{h}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}\left( {\dfrac{{\sin \left( {\dfrac{h}{2}} \right)}}{{\dfrac{h}{2}}}} \right)} \right\}\,
 =  - \cos ec x \,\cot x
(11)f(x) = \cot x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cot \left( {x + h} \right) - \cot x}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin x\cos \left( {x + h} \right) - \cos x\sin \left( {x + h} \right)}}{{h\sin x\sin \left( {x + h} \right)}}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \sin h}}{{h\sin x\sin \left( {x + h} \right)}}\,\,\,\,\,\left\{ \begin{array}{l}  {\rm{Notice\, how\, the\, numerator}}\\  {\rm{ was \,simplified}}  \end{array} \right\}
 =  - \cos e{c^2}x
(12)f(x) = {e^x}
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{x + h}} - {e^x}}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^x} \cdot {e^h} - {e^x}}}{h}
 = {e^x}\mathop {\lim }\limits_{h \to 0} \dfrac{{{e^h} - 1}}{h}
 = {e^x}
(13)f(x) = {a^x}
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{{a^{x + h}} - {a^x}}}{h}
 = {a^x} \cdot \mathop {\lim }\limits_{h \to 0} \dfrac{{{a^h} - 1}}{h}
 = {a^x}\ln a
(14)f(x) = \ln x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
= \mathop {\lim }\limits_{h \to 0} \dfrac{{\ln \left( {x + h} \right) - \ln x}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{\ln \left( {1 + \dfrac{h}{x}} \right)}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{\ln \left( {1 + \dfrac{h}{x}} \right)}}{{x \cdot \dfrac{h}{x}}}
 = \dfrac{1}{x}\,\,\,\,\,\,\,\,\left\{ {{\rm{Because}}\mathop {\lim }\limits_{\theta  \to 0} \left( {\dfrac{{\ln \left( {1 + \theta } \right)}}{\theta }} \right) = 1} \right\}
(15)f(x) = {\log _a}x
Since {\log _a}x can be written as \dfrac{{\ln x}}{{\ln a}},\dfrac{{d\left( {{{\log }_a}x} \right)}}{{dx}} will be \dfrac{1}{{\ln a}}\dfrac{{d\ln \left( x \right)}}{{dx}} {because \dfrac{1}{{\ln a}}is a constant so it can be taken outside the differentiation operator; we will prove the validity of this step later}.
Therefore,
\dfrac{d}{{dx}}\left( {{{\log }_a}x} \right) = \dfrac{1}{{x\ln a}}
(16)f(x) = {\sin ^{ - 1}}x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^{ - 1}}\left( {x + h} \right) - {{\sin }^{ - 1}}x}}{h}
\left\{ \begin{array}{l}  {\rm{The\, numerator\, can\, be\, simplified\, as\, follows: }}\\  {\rm{Let\, us\, consider\, a\, general\, expression\, si}}{{\rm{n}}^{{\rm{-1}}}}x\, -\, {\sin ^{ - 1}}y{\rm{. }}\\  {\rm{Let\, si}}{{\rm{n}}^{{\rm{-1}}}}x = p\,{\rm{and}}\,{\sin ^{ - 1}}y = q,\,\,{\rm{so \, that \,}}x = \sin p\,{\rm{and}}\,y = \sin q\,\,\\  {\rm{Now}},\,\,\sin \left( {p - q} \right) = \sin p\cos q - \cos p\sin q\\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,x\sqrt {1 - {y^2}}  - y\sqrt {1 - {x^2}} \\  {\rm{Therefore}},\\  p - q = {\sin ^{ - 1}}x - {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left( {x\sqrt {1 - {y^2}}  - y\sqrt {1 - {x^2}} } \right)  \end{array} \right\}
We use this relation now to simplify the numerator:
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^{ - 1}}\left\{ {\left( {x + h} \right)\sqrt {1 - {x^2}}  - x\sqrt {1 - {{\left( {x + h} \right)}^2}} } \right\}}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^{ - 1}}y}}{h}  \,\,\,\,\left\{ \begin{array}{l}  {\rm{we \,replaced\, the\, large }}\\  {\rm{argument\, of\, si}}{{\rm{n}}^{{\rm{-1}}}}\,{\rm{ by }}y{\rm{ }}  \end{array} \right\}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^{ - 1}}y}}{y} \cdot \dfrac{y}{h}
Notice that as h \to 0,\,\,y \to 0\,\,\,{\rm{so}}\,\,\dfrac{{{{\sin }^{ - 1}}y}}{y} \to 1
Also,
\mathop {\lim }\limits_{h \to 0} \dfrac{y}{h} = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {x + h} \right)\sqrt {1 - {x^2}}  - x\sqrt {1 - {{\left( {x + h} \right)}^2}} }}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\left( {x + h} \right)}^2}\left( {1 - {x^2}} \right) - {x^2}\left( {1 - {{\left( {x + h} \right)}^2}} \right)}}{{h\left\{ {\left( {x + h} \right)\sqrt {1 - {x^2}}  + x\sqrt {1 - {{\left( {x + h} \right)}^2}} } \right\}}}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{h^2} + 2xh}}{{h\left\{ {\left( {x + h} \right)\sqrt {1 - {x^2}}  + x\sqrt {1 - {{\left( {x + h} \right)}^2}} } \right\}}}
 = \dfrac{{2x}}{{2x\sqrt {1 - {x^2}} }}
 = \dfrac{1}{{\sqrt {1 - {x^2}} }}
Therefore,
f'\left( x \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}
(17)f(x) = {\cos ^{ - 1}}x
Notice that {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}
Thus,
\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right) = \dfrac{{d\left( {\dfrac{\pi }{2}} \right)}}{{dx}} = 0
 \Rightarrow   \dfrac{{d\left( {{{\cos }^{ - 1}}x} \right)}}{{dx}}
= \dfrac{{ - d\left( {{{\sin }^{ - 1}}x} \right)}}{{dx}}
 = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}
Note that for the last step, we have used the fact that differentiation operation is distributive over addition, i.e. (f + g)' = f' + g'. We will justify this later.
(18)f(x) = {\tan ^{ - 1}}x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\tan }^{ - 1}}\left( {x + h} \right) - {{\tan }^{ - 1}}x}}{h}
 = \mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{{{\tan }^{ - 1}}\dfrac{h}{{1 + x\left( {x + h} \right)}}}}{h}} \right\} \cdot\,\,\,\,\left( \begin{array}{l}  {\rm{we \,used}}\,{\tan ^{ - 1}}A - {\tan ^{ - 1}}B\\   = {\tan ^{ - 1}}\left( {\dfrac{{A - B}}{{1 + AB}}} \right);{\rm{verify \,this}}  \end{array} \right)
= \mathop {\lim }\limits_{h \to 0} \left\{ {\dfrac{{{{\tan }^{ - 1}}\dfrac{h}{{1 + x(x + h)}}}}{{\dfrac{h}{{1 + x(x + h)}}}}} \right\}.\dfrac{1}{{1 + x(x + h)}}
 = \dfrac{1}{{1 + {x^2}}}
(19)f(x) = \cos e{c^{ - 1}}x
f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos {\rm{e}}{{\rm{c}}^{ - 1}}\left( {x + h} \right) - \cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^{ - 1}}\left( {\dfrac{1}{{x + h}}} \right) - {{\sin }^{ - 1}}\left( {\dfrac{1}{x}} \right)}}{h}\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}  {\rm{Notice \,that}}\\  \cos {\rm{e}}{{\rm{c}}^{ - 1}}\theta  = {\sin ^{ - 1}}\dfrac{1}{\theta }  \end{array} \right)
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^{ - 1}}\left\{ {\left( {\dfrac{1}{{x + h}}} \right)\sqrt {1 - {{\left( {\dfrac{1}{x}} \right)}^2}}  - \left( {\dfrac{1}{x}} \right)\sqrt {1 - {{\left( {\dfrac{1}{{x + h}}} \right)}^2}} } \right\}}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^{ - 1}}\left\{ {\dfrac{{\sqrt {{x^2} - 1} }}{{\left| x \right|\left( {x + h} \right)}} - \dfrac{{\sqrt {{{\left( {x + h} \right)}^2} - 1} }}{{x\left| {x + h} \right|}}} \right\}}}{h}
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^{ - 1}}y}}{h}  \,\,\,\,\left( \begin{array}{l}  {\rm{where }}\,y\,{\rm{is \,the\, argument\, of }}\,{\sin ^{ - 1}};\\  {\rm{Note\, that}}\,\,y \to 0\,\,{\rm{as}}\,\,h \to 0  \end{array} \right)
 = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{\sin }^{ - 1}}y}}{y} \cdot \dfrac{y}{h}
Now, it can easily be verified by rationalization that
\mathop {\lim }\limits_{h \to 0} \dfrac{y}{h} = \dfrac{{ - 1}}{{\left| x \right|\sqrt {{x^2} - 1} }}
Therefore,
f'\left( x \right) = \dfrac{{ - 1}}{{\left| x \right|\sqrt {{x^2} - 1} }}
(20)f(x) = {\sec ^{ - 1}}x
Notice that {\sec ^{ - 1}}x + {\rm{cose}}{{\rm{c}}^{ - 1}}x = \dfrac{\pi }{2}
Therefore,
\dfrac{{d\left( {{{\sec }^{ - 1}}x} \right)}}{{dx}}
 = \dfrac{{ - d\left( {{\rm{cose}}{{\rm{c}}^{ - 1}}x} \right)}}{{dx}}
 = \dfrac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}
(21)f(x) = {\cot ^{ - 1}}x
Notice again that
{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}
Therefore, as in the earlier cases,
\dfrac{{d\left( {{{\cot }^{ - 1}}x} \right)}}{{dx}} = \dfrac{{ - d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}} = \dfrac{{ - 1}}{{1 + {x^2}}}
It would be of help to you to get used to these differentiation on formulae as soon as possible, since they will be widely used subsequently.
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