Differentiation Of Exponential Logarithmic And Inverse Trignometric Functions

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12. \(\boxed{f\left( x \right) = {e^x}}\) :

\[\begin{align}f'\left( x \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{x + h}} - {e^x}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^x} \cdot {e^h} - {e^x}}}{h}\\& = {e^x}\mathop {\lim }\limits_{h \to 0} \frac{{{e^h} - 1}}{h}\\ &= {e^x}\end{align}\]

13. \(\boxed{f\left( x \right) = {a^x}}\) :

\[\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{a^{x + h}} - {a^x}}}{h}\\&= {a^x} \cdot \mathop {\lim }\limits_{h \to 0} \frac{{{a^h} - 1}}{h}\\ &= {a^x}\ln a\end{align}\]

14. \(\boxed{f\left( x \right) = \ln x}\) :

\[\begin{align}f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} & = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {x + h} \right) - \ln x}}{h}\\& = \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{x}} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\ln \left( {1 + \frac{h}{x}} \right)}}{{x \cdot \frac{h}{x}}}\\& = \frac{1}{x}\left\{ {{\rm{Because}}\mathop {\lim }\limits_{\theta \to 0} \left( {\frac{{\ln \left( {1 + \theta } \right)}}{\theta }} \right) = 1} \right\}\end{align}\]

15. \(\boxed{f\left( x \right) = {{\log }_a}x}\) :

Since \({\log _a}x\) can be written as \(\frac{{\ln x}}{{\ln a}},\frac{{d\left( {{{\log }_a}x} \right)}}{{dx}}\)  will be \(\frac{1}{{\ln a}}\frac{{d\ln \left( x \right)}}{{dx}}\) {because \(\frac{1}{{\ln a}}\) is a constant so it can be taken outside the differentiation operator; we will prove the validity of this step later}.

Therefore, \(\frac{d}{{dx}}\left( {{{\log }_a}x} \right) = \frac{1}{{x\ln a}}\)

16. \(\boxed{f\left( x \right) = {{\sin }^{ - 1}}x}\):

\(f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}\left( {x + h} \right) - {{\sin }^{ - 1}}x}}{h}\)

\[\left\{ \begin{align}&{\rm{The\;numerator\;can\;be\;simplified\;as\;follows:\;}}\\&{\rm{Let\;us\;consider\;a\;general\;expression\;si}}{{\rm{n}}^{{\rm{-1}}}}x - {\sin ^{ - 1}}y{\rm{. }}\\&{\rm{Let \;si}}{{\rm{n}}^{{\rm{-1}}}}x = p\,{\rm{and}}\,{\sin ^{ - 1}}y = q,\,\,{\rm{so that }}x = \sin p\,{\rm{and}}\,y = \sin q\,\,\\&{\rm{Now}},\,\,\sin \left( {p - q} \right) = \sin p\cos q - \cos p\sin q\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,x\sqrt {1 - {y^2}} - y\sqrt {1 - {x^2}} \\&{\rm{Therefore}},\\&p - q = {\sin ^{ - 1}}x - {\sin ^{ - 1}}y = {\sin ^{ - 1}}\left( {x\sqrt {1 - {y^2}} - y\sqrt {1 - {x^2}} } \right)\end{align} \right\}\]

We use this relation now to simplify the numerator:

\[\begin{align}f'\left( x \right)&= \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}\left\{ {\left( {x + h} \right)\sqrt {1 - {x^2}} - x\sqrt {1 - {{\left( {x + h} \right)}^2}} } \right\}}}{h}\\& = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}y}}{h} \,\,\,\,\left\{ \begin{array}{l}{\rm{we\;replaced\;the\;large\;}}\\{\rm{argument\;of\;si}}{{\rm{n}}^{{\rm{-1}}}}{\rm{ by\; }}y{\rm{ }}\end{array} \right\}\\& = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}y}}{y} \cdot \frac{y}{h}\end{align}\]

Notice that as \(h \to 0,\,\,y \to 0\,\,\,{\rm{so}}\,\,\frac{{{{\sin }^{ - 1}}y}}{y} \to 1\)

\(\begin{align}\text{Also, }\qquad \qquad\qquad \mathop {\lim }\limits_{h \to 0} \frac{y}{h} &= \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h} \right)\sqrt {1 - {x^2}} - x\sqrt {1 - {{\left( {x + h} \right)}^2}} }}{h}\\& = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^2}\left( {1 - {x^2}} \right) - {x^2}\left( {1 - {{\left( {x + h} \right)}^2}} \right)}}{{h\left\{ {\left( {x + h} \right)\sqrt {1 - {x^2}} + x\sqrt {1 - {{\left( {x + h} \right)}^2}} } \right\}}}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + 2xh}}{{h\left\{ {\left( {x + h} \right)\sqrt {1 - {x^2}} + x\sqrt {1 - {{\left( {x + h} \right)}^2}} } \right\}}}\\& = \frac{{2x}}{{2x\sqrt {1 - {x^2}} }}\\ &= \frac{1}{{\sqrt {1 - {x^2}} }}\end{align}\)

Therefore,

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

17. \(\boxed{f\left( x \right) = {{\cos }^{ - 1}}x}\) :

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

\( \begin{align}\text{Thus,}\qquad \qquad \qquad &\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right)  = \frac{{d\left( {\frac{\pi }{2}} \right)}}{{dx}} = 0\\&\Rightarrow \frac{{d\left( {{{\cos }^{ - 1}}x} \right)}}{{dx}} = \frac{{ - d\left( {{{\sin }^{ - 1}}x} \right)}}{{dx}} = \frac{{ - 1}}{{\sqrt {1 - {x^2}} }} \end{align}\)

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. \(\boxed{f\left( x \right) = {{\tan }^{ - 1}}x}\) :

\[\begin{align}f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\tan }^{ - 1}}\left( {x + h} \right) - {{\tan }^{ - 1}}x}}{h}\end{align}\]

\[\begin{align}& = \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{{{\tan }^{ - 1}}\frac{h}{{1 + x\left( {x + h} \right)}}}}{{\frac{h}{{1 + x\left( {x + h} \right)}}}}} \right\} \cdot \frac{1}{{1 + x\left( {x + h} \right)}}\\ &= \frac{1}{{1 + {x^2}}}\end{align}\]

19. \(\boxed{f\left( x \right) = \cos {\text{e}}{{\text{c}}^{ - 1}}x}\) :

\[\begin{align}f'\left( x \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{\cos {\rm{e}}{{\rm{c}}^{ - 1}}\left( {x + h} \right) - \cos {\rm{e}}{{\rm{c}}^{ - 1}}x}}{h}\\& = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}\left( {\frac{1}{{x + h}}} \right) - {{\sin }^{ - 1}}\left( {\frac{1}{x}} \right)}}{h}\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}{\rm{Notice that}}\\\cos {\rm{e}}{{\rm{c}}^{ - 1}}\theta = {\sin ^{ - 1}}\frac{1}{\theta }\end{array} \right)\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}\left\{ {\left( {\frac{1}{{x + h}}} \right)\sqrt {1 - {{\left( {\frac{1}{x}} \right)}^2}} - \left( {\frac{1}{x}} \right)\sqrt {1 - {{\left( {\frac{1}{{x + h}}} \right)}^2}} } \right\}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}\left\{ {\frac{{\sqrt {{x^2} - 1} }}{{\left| x \right|\left( {x + h} \right)}} - \frac{{\sqrt {{{\left( {x + h} \right)}^2} - 1} }}{{x\left| {x + h} \right|}}} \right\}}}{h}\\& = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}y}}{h} \,\,\,\,\left( \begin{array}{l}{\rm{where\;}}y\,{\rm{is\;the\;argument\;of\;}}{\sin ^{ - 1}};\\{\rm{Note\;that}}\,\,y \to 0\,\,{\rm{\;as}}\,\,h \to 0 \end{array} \right)\\& = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^{ - 1}}y}}{y} \cdot \frac{y}{h}\end{align}\]

Now, it can easily be verified by rationalization that

\[\mathop {\lim }\limits_{h \to 0} \frac{y}{h} = \frac{{ - 1}}{{\left| x \right|\sqrt {{x^2} - 1} }}\]

Therefore,

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

20. \(\boxed{f\left( x \right) = {{\sec }^{ - 1}}x}\) :

Notice that \({\sec ^{ - 1}}x + {\rm{cose}}{{\rm{c}}^{ - 1}}x = \frac{\pi }{2}\)

Therefore, \(\frac{{d\left( {{{\sec }^{ - 1}}x} \right)}}{{dx}} = \frac{{ - d\left( {{\rm{cose}}{{\rm{c}}^{ - 1}}x} \right)}}{{dx}} = \frac{1}{{\left| x \right|\sqrt {{x^2} - 1} }}\)

21. \(\boxed{f\left( x \right) = {{\cot }^{ - 1}}x}\) :

Notice again that \(\begin{align}{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}\end{align}\)

Therefore, as in the earlier cases,

\[\begin{align}\frac{{d\left( {{{\cot }^{ - 1}}x} \right)}}{{dx}} = \frac{{ - d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}} = \frac{{ - 1}}{{1 + {x^2}}}\end{align}\]

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.