mathematics-class-12-unit-06-chapter-03-derivatives-l-3-12-gioozpmf-rg

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Example 1 </primary>
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- We saw that (i) $\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}$

- (ii) $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$

- We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$

- $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$

- $\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}$

- $ \therefore \quad \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x $

- $ \Rightarrow \frac{d}{d x}\left(\cos ^{-1} x\right)=0-\frac{d}{d x}\left(\sin ^{-1} x\right)=-\frac{1}{\sqrt{1-x^2}} $

- similarly $ \frac{d}{d x}\left(\cot ^{-1} x\right)=-\frac{d}{d x}\left(\tan ^{-1} x\right)=-\frac{1}{1+x^2}$

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<footer style = "font-size: 18px"> $\rightarrow$ Derivatives lecture - 3 $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Example 2 $\rightarrow$ Example 2 </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Example 2 </primary>
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- Let's calculate $\frac{d}{d x}\left(\sec ^{-1} x\right)$.

- Let $y=\sec ^{-1} x$ the $x=\sec y$    
- Diff. w.r.t. $x$, we got

- $1=\frac{d}{d x}(\sec y)  =\frac{d}{d y}(\sec y) \cdot \frac{d y}{d x} \text { (by Chain rule) }$

- $ =\sec y \tan y \frac{d y}{d x}$

- $\Rightarrow \frac{d y}{d x}  =\frac{1}{\sec y \tan y}$

- Since, $\sec y=x, \quad \tan ^2 y=\sec ^2 y-1=x^2-1$    
- $\Rightarrow \tan y= \pm \sqrt{x^2-1}$

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<footer style = "font-size: 18px">Derivatives lecture - 3 $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Example 2 $\rightarrow$ Example 2 </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Example 2 </primary>
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- $\sec ^{-1} x \in\left[0, \frac{\pi}{2}\right)$ if $x \in[1, \infty) \sec ^{-1} x \in\left(\frac{\pi}{2}, \pi\right] \text { if } x \in(-\infty,-1]$     
- ( $\sec ^{-1} x$ is not defined if $-1<x<1$ )      
- $  \therefore \quad y=\sec ^{-1} x \in\left[0, \frac{\pi}{2}\right) \text { if } x \in[1, \infty) $     
- $ \in\left(\frac{\pi}{2}, \pi\right] \text { if } x \in(-\infty,-1] $       
- $ \Rightarrow \quad \tan y \geqslant 0 \text { if } x \in[1, \infty) $      
- $ \leqslant 0 \text { if } x \in(-\infty,-1] $       
- $ \therefore \quad \tan y= \begin{cases}\sqrt{x^2-1} & \text { if } x \geqslant 1 \\\ -\sqrt{x^2-1}  \text { if } x \leqslant-1\end{cases} $    
- $\tan \left(\sec ^{-1} x\right)$

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<footer style = "font-size: 18px">Example 1 $\rightarrow$ Example 2 $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Example 2 $\rightarrow$ Implicit Differentiation </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Example 2 </primary>
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- $\therefore \quad \frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{\sec y \tan y}= \begin{cases}\frac{1}{x \sqrt{x^2-1}}  \text { if } x \geqslant 1 \\\ \frac{1}{-x \sqrt{x^2-1}}\text { if } x \leqslant 1\end{cases}$

- Since $|x|=\{\begin{cases}{}x \text { if } x \geqslant 0 \\\ -x  \text { if } x<0\end{cases}}$,

- $\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}} \quad \text { if }|x| \geqslant 1$

- Hence, $\frac{d}{d x}\left(\operatorname{cosec}^{-1} x\right)=\frac{-1}{|x| \sqrt{x^2-1}}$ if $|x| \geqslant 1$

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<footer style = "font-size: 18px">Example 2 $\rightarrow$ Example 2 $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Implicit Differentiation $\rightarrow$ Example 3 </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Example 3 </primary>
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- $ y+\sin y=x $    
- $\Rightarrow  \frac{d}{d x}(y+\sin y)=\frac{d}{d x}(x) $      
- $\Rightarrow  \frac{d y}{d x}+\frac{d}{d x}(\sin y)=1 $        
- $\Rightarrow  \frac{d y}{d x}+\frac{d}{d y}(\sin y) \frac{d y}{d x}=1 \quad \text { (By Chain rule) } $         
- $\Rightarrow \quad  \frac{d y}{d x}[1+\cos y]=1 $     
- $\Rightarrow  \frac{d y}{d x}=\frac{1}{1+\cos y} \text {, provided cosy } \neq-1 .$

- Note that by doing implicit differentiation, we need not get $\frac{d y}{d x}$ as a function of $x$ only.
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<footer style = "font-size: 18px">Example 2 $\rightarrow$ Implicit Differentiation $\rightarrow$ <span style = "color:blue"> Example 3 </span> $\rightarrow$ Exponential function $\rightarrow$ Properties of exp (x) </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Exponential function </primary>
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- We define       
- $ \exp (x)  =1+\frac{x}{1 !}+\frac{x^2}{2 !}+\cdots+\frac{x^n}{n !}+\cdots \cdot $       
- $   =\sum_{n=0}^{\infty} \frac{x^n}{n !} $

- = $ \lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{x^k}{k !} $

- Fact: The above series converges (to a finite real number) for every $x \in \mathbb{R}$.

- Thus $\exp (x)$ is defined for every real no $x$.
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<footer style = "font-size: 18px">Implicit Differentiation $\rightarrow$ Example 3 $\rightarrow$ <span style = "color:blue"> Exponential function </span> $\rightarrow$ Properties of exp (x) $\rightarrow$ Properties of exp(x) </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Properties of exp (x) </primary>
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- Further, $\exp (x)$ satisfies the following properties:
- (1) $\exp (0)=1$      
- (2) $\exp (x)$ is an increasing function of $x$ ie. $\quad x_1<x_2 \Rightarrow \exp \left(x_1\right)<\exp \left(x_2\right)$.      
- (3) Domain $=\mathbb{R}$         
- (4) Range of exponential for $=\mathbb{R}^{+}=(0, \infty)$ ( $\exp (x)$ is never zero or negative)        
- (5) $\lim _{x \rightarrow+\infty} \exp (x)=+\infty$        
- (6) $\lim _{x \rightarrow-\infty} \exp (x)=0$.       
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<footer style = "font-size: 18px">Example 3 $\rightarrow$ Exponential function $\rightarrow$ <span style = "color:blue"> Properties of exp (x) </span> $\rightarrow$ Properties of exp(x) $\rightarrow$ Graph of exp(x) </footer>

<h3 id="success-stylefont-size25px-derivatives-l-3-success"><Success style="font-size:25px"> Derivatives L-3 </Success></h3>
<h3 id="primary-stylefont-size24px-properties-of-expx-primary"><primary style="font-size:24px"> Properties of exp(x) </primary></h3>
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<p>(7) $\exp (x+y)=\exp (x) \exp (y)$</p>
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<p>(Compare with $a^{m+n}=a^m \cdot a^n$ if $m, n \in \mathbb{N}$ ) Using (7) and (1),  we get</p>
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<p>$ \exp (-x)  =\frac{1}{\exp (x)} $</p>
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<p>$ (\because 1=\exp (0)  =\exp (x+(-x))=\exp (x) \exp (-x)) $</p>
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<p>So, (6) follows from (5).</p>
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<footer style = "font-size: 18px">Exponential function $\rightarrow$ Properties of exp (x) $\rightarrow$ <span style = "color:blue"> Properties of exp(x) </span> $\rightarrow$ Graph of exp(x) $\rightarrow$ Expansion of e </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Expansion of e </primary>
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- Putting $x=1$, we get
- $\begin{aligned}\exp (1) & =1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{n !}+\cdots \\& =\sum_{k=0}^{\infty} \frac{1}{k !}\end{aligned}$

- We know that $\exp (1)$ in a real, and we denote this by $e$ (Euler's constant).
- So, $e=\exp (1)=\sum_{k=0}^{\infty} \frac{1}{k !}$

- We can show that $2<e<3$
- (In fact, $e \approx 2.718 \ldots$ )
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<footer style = "font-size: 18px">Properties of exp(x) $\rightarrow$ Graph of exp(x) $\rightarrow$ <span style = "color:blue"> Expansion of e </span> $\rightarrow$ 2< e< 3 $\rightarrow$ 2< e< 3 </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> 2< e< 3 </primary>
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- $ e=1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots>1+\frac{1}{1 !}=2 .$

- Also, $e=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots$
- $<1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots \cdot \quad \quad $  (because $\frac{1}{(n+1) !}<\frac{1}{2^n}$ for $n \geqslant 2$ )

- Recall: Geometric series $ \quad\quad a+a r+a r^2+\cdots=\frac{a}{1-r} \text { if }|\gamma|<1 $

- Putting $a=1, r = \frac{1}{2}$ gives
- $ 1+\frac{1}{2}+\frac{1}{2^2}+\cdots=\frac{1}{1-x_2}=2$

- $\therefore \quad e<1+2=3 $

- Fact: It can be proved that $e$ is an irrational number. Also,
- $\begin{aligned}& \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n=e . \\& \lim _{x \rightarrow 0}(1+x)^{1 / x}=e .\end{aligned} $

- Let's calculate $\lim _{h \rightarrow 0} \frac{\exp (h)-1}{h}$

- Notation: We write $\exp (x)$ as $e^x$

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<footer style = "font-size: 18px">Graph of exp(x) $\rightarrow$ Expansion of e $\rightarrow$ <span style = "color:blue"> 2< e< 3 </span> $\rightarrow$ 2< e< 3 $\rightarrow$ Derivation of exp(x) </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> 2< e< 3 </primary>
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- $ e^h=\exp (h)=1+\frac{h}{1 !}+\frac{h^2}{2 !}+\cdots+\frac{h^n}{n !}+\cdots $      
- $ \Rightarrow \quad e^h-1=h+\frac{h^2}{2 !}+\frac{h^3}{3 !}+\cdots+\frac{h^n}{n !}+\cdots $       
- $ \Rightarrow \quad \frac{e^h-1}{h}=1+\frac{h}{2 !}+\frac{h^2}{3 !}+\cdots+\frac{h^{n !}}{n !}+\cdots \text { for } h \neq 0 . $       
- $ \lim _{h \rightarrow 0} \frac{e^h-1}{h}=1 $

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<footer style = "font-size: 18px">Expansion of e $\rightarrow$ 2< e< 3 $\rightarrow$ <span style = "color:blue"> 2< e< 3 </span> $\rightarrow$ Derivation of exp(x) $\rightarrow$ Examples </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Derivation of exp(x) </primary>
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- Derivative of $e^x=\exp (x)$.     
- $f(x)=e^x$    
- $ f^{\prime}(x)  =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $        
- $ =\lim _{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\lim _{h \rightarrow 0} \frac{e^x e^h-e^x}{h} $     
- $ =e^x \lim _{h \rightarrow 0}\left(\frac{e^h-1}{h}\right)=e^x .$

- So, $\frac{d}{d x}\left(e^x\right)=e^x$

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<footer style = "font-size: 18px">2< e< 3 $\rightarrow$ 2< e< 3 $\rightarrow$ <span style = "color:blue"> Derivation of exp(x) </span> $\rightarrow$ Examples $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Derivatives L-3 </Success>
### <primary style="font-size:24px"> Examples </primary>
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- (1). $\frac{d}{d x}\left(e^{5 x}\right)=\frac{d}{d(5 x)}\left(e^{5 x}\right) \cdot \frac{d}{d x}(5 x)=e^{5 x} \cdot 5$       
- (2). $\frac{d}{d x}\left(e^{x^2}\right)=e^{x^2} \cdot \frac{d}{d x}\left(x^2\right)=2 x e^{x^2}$.            
- (3). $\frac{d}{d x} \tan ^{-1}\left(e^x\right)=\frac{1}{1+\left(e^x\right)^2} \cdot \frac{d}{d x}\left(e^{-x}\right)=\frac{e^x}{1+e^{2 x}}$

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<footer style = "font-size: 18px">2< e< 3 $\rightarrow$ Derivation of exp(x) $\rightarrow$ <span style = "color:blue"> Examples </span> $\rightarrow$ Thank you $\rightarrow$  </footer>