In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

Differentiation Of Exponential Logarithmic And Inverse Trignometric Functions

Go back to  'LCD'

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.

Download SOLVED Practice Questions of Differentiation Of Exponential Logarithmic And Inverse Trignometric Functions for FREE
Limits
grade 11 | Questions Set 1
Limits
grade 11 | Answers Set 1
Limits
grade 11 | Questions Set 2
Limits
grade 11 | Answers Set 2