# Differentiation Of Polynomial And Trigonometric Functions

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## 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. $$\boxed{f\left( x \right) = k}:$$

\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{{k - k}}{h} = 0\end{align}

This is intuitively true also since the graph for a constant function is a horizontal line.

2. $$\boxed{f\left( x \right) = x}:$$

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

This corresponds to the fact that the line $$f(x) = x$$ is inclined at 45º to the x–axis (and tan 45º is 1).

3. $$\boxed{f\left( x \right) = mx + c}:$$

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

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. $$\boxed{f\left( x \right) = {x^2}}:$$

$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^2} - {x^2}}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + 2xh}}{h} = 2x$

We have already obtained this result earlier.

5. $$\boxed{f\left( x \right) = {x^n}}:$$

\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{{{{\left( {x + h} \right)}^n} - {x^n}}}{h}\\ &=\underset{h\to 0}{\mathop{\lim }}\,\left\{ \frac{{{x}^{n}}{{\left( 1+\frac{h}{x} \right)}^{n}}-{{x}^{n}}}{h} \right\} \\ &={{x}^{n}}\underset{h\to 0}{\mathop{\lim }}\,\left\{ \frac{{{\left( 1+\frac{h}{x} \right)}^{n}}-1}{h} \right\} \\ &={{x}^{n}}\underset{h\to 0}{\mathop{\lim }}\,\frac{\left\{ 1+\frac{nh}{x}+n\frac{\left( n-1 \right)}{2!}\frac{{{h}^{2}}}{{{x}^{2}}}+.... \right\}-1}{h}~~~~~~~~\left[ \text{By the binomial expansion} \right] \\& ={{x}^{n}}\cdot \frac{n}{x}=n{{x}^{n-1}} \end{align}

So, for example, \begin{align}\frac{d({{x}^{2}})}{dx}=2.{{x}^{2-1}}=2x~\text{ }and~\text{ }\frac{d({{x}^{3}})}{dx}=3.{{x}^{3-1}}=3{{x}^{2}}\end{align} and so on.

6. $$\boxed{f\left( x \right) = \sin x}:$$

\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{{\sin \left( {x + h} \right) - \sin x}}{h}\\& = \mathop {\lim }\limits_{h \to 0} \frac{{2\cos \left( {x + \frac{h}{2}} \right)\sin \left( {\frac{h}{2}} \right)}}{h}\\& = \mathop {\lim }\limits_{h \to 0} \left\{ {\cos \left( {x + \frac{h}{2}} \right) \cdot \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right\}\\ &= {\rm{ }}cosx.\end{align}

7. $$\boxed{f\left( x \right) = \cos x}$$ :

\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 \left( {x + h} \right) - \cos x}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{ - 2\sin \left( {x + \frac{h}{2}} \right)\sin \frac{h}{2}}}{h}} \right\}\\& = \mathop {\lim }\limits_{h \to 0} \left\{ { - \sin \left( {x + \frac{h}{2}} \right) \cdot \frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right\}\\&= {\rm{ }}-{\rm{ }}sinx.\end{align}

8. $$\boxed{f\left( x \right) = \tan x}$$:

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

9. $$\boxed{f\left( x \right) = \sec x}:$$

\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{{\sec \left( {x + h} \right) - \sec x}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{\cos x - \cos \left( {x + h} \right)}}{{h\cos x\cos \left( {x + h} \right)}}\\ &= \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{2\sin \left( {x + \frac{h}{2}} \right)\sin \frac{h}{2}}}{{h\cos x\cos \left( {x + h} \right)}}} \right\}\\ &= \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{\sin \left( {x + \frac{h}{2}} \right)}}{{\cos x\cos \left( {x + h} \right)}}\left( {\frac{{\sin \frac{h}{2}}}{{\frac{h}{2}}}} \right)} \right\}\\ &= \,{\rm{Sec}}\,x\,{\rm{tan}}\,x{\rm{ }} \end{align}

10. $$\boxed{f\left( x \right) = \cos {\text{ec}}\,x}$$ :

\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{{{\mathop{\rm cosec}\nolimits} \left( {x + h} \right) - {\mathop{\rm cosec}\nolimits} x}}{h}\\& = \mathop {\lim }\limits_{h \to 0} \frac{{\sin x - \sin \left( {x + h} \right)}}{{h\sin x\sin \left( {x + h} \right)}}\\\ &= \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{ - 2\cos \left( {x + \frac{h}{2}} \right)\sin \frac{h}{2}}}{{h\sin x\sin \left( {x + h} \right)}}} \right\}\\ &= \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{ - \cos \left( {x + \frac{h}{2}} \right)}}{{\sin x\sin \left( {x + h} \right)}}\left( {\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\frac{h}{2}}}} \right)} \right\}\,\\ &= \text{cosec }x \; \text{cot } x\end{align}

11. $$\boxed{f\left( x \right) = \cot x}$$ :

\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{{\cot \left( {x + h} \right) - \cot x}}{h}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{\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} \frac{{ - \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\}\\ &= {\rm{ }}-{\rm{ }}\text{cosec}^2x\end{align}