mathematics-class-12-unit-06-chapter-04-derivatives

### <Success style="font-size:25px"> Derivatives L-4 </Success>
### <primary style="font-size:24px"> Some examples </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 \quad \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}} $

- $ \text { Il y, } \quad \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-4 $\rightarrow$ <span style = "color:blue"> Some examples </span> $\rightarrow$ $\frac{d}{d x}\left(\sec ^{-1} x\right)$ $\rightarrow$ $\frac{d}{d x}\left(\sec ^{-1} x\right)$ </footer>

### <Success style="font-size:25px"> Derivatives L-4 </Success>
### <primary style="font-size:24px"> $\frac{d}{d x}\left(\sec ^{-1} x\right)$ </primary>
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- Let's calculate $\frac{d}{d x}\left(\sec ^{-1} x\right)$.

- Let $y=\sec ^{-1} x$ Then $x=\sec y$

- Diff. w.r.t. $x$, we get

- $1=\frac{d}{d x}(\sec y)  =\frac{d}{d y}(\sec y) \cdot \frac{d y}{d x}$

- (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-4 $\rightarrow$ Some examples $\rightarrow$ <span style = "color:blue"> $\frac{d}{d x}\left(\sec ^{-1} x\right)$ </span> $\rightarrow$ $\frac{d}{d x}\left(\sec ^{-1} x\right)$ $\rightarrow$ Implicit Differentiation </footer>

### <Success style="font-size:25px"> Derivatives L-4 </Success>
### <primary style="font-size:24px"> $\frac{d}{d x}\left(\sec ^{-1} x\right)$ </primary>
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- Recall : $\sec ^{-1} x \in\left(0, \frac{\pi}{2}\right)$ if $x \in(1, \infty)$

- $2 \sec ^{-1} x \in\left(\frac{\pi}{2}, \pi\right) \text { if } x \in(-\infty,-1)$

- ( $\sec ^{-1} x$ is not defined if $-1<x<1$ )

- $ \therefore 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 \quad \text { if } x \in(1, \infty) $              
- $ ~ \qquad\leqslant 0 \quad \text { if } x \in(-\infty,-1) $

- $\tan y= \begin{cases}\sqrt{x^2-1}  \text { if } x \geqslant 1 \\\\ \sqrt{x^2-1}  \text { if } x \leqslant-1\end{cases}$

- $\tan \left(\sec ^{-1} x\right)$

- $\therefore \quad \frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{\sec y \tan y}= \begin{cases}\frac{1}{x \sqrt{x^2-1}} & \text { if } x \geqslant 1, \\ \qquad \frac{1}{-x \sqrt{x^2-1}} & \text { if } x \leqslant 1\end{cases}$

- $\text { Since }|x|= x \text { if } x \geqslant 0 \quad or \quad -x\text { if } x<0$

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<footer style = "font-size: 18px">Some examples $\rightarrow$ $\frac{d}{d x}\left(\sec ^{-1} x\right)$ $\rightarrow$ <span style = "color:blue"> $\frac{d}{d x}\left(\sec ^{-1} x\right)$ </span> $\rightarrow$ Implicit Differentiation $\rightarrow$ Exponential Function </footer>

### <Success style="font-size:25px"> Derivatives L-4 </Success>
### <primary style="font-size:24px"> Implicit Differentiation </primary>
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- Sometimes we have am equation relating $y$ to $x$ , but it is difficult to write explicitly $y$ as a fr. of $x$'

- <ins>For Example:</ins> $ ~ ~ y + \sin  = x $

- Find $=\frac{dy}{dx}$

- Here instead of trying to write $y$ as a function of $x$, we will do implicit differentiation as follows.

- y + sin y $=x$

- $\Rightarrow \frac{d}{dx} =(y + \sin y) = \frac{d}{dx}(x)$$\quad$
- $\Rightarrow\frac{dy}{dx} + \frac{d}{dx}(\sin y)=1$

- $\Rightarrow\frac{dy}{dx} + \frac{d}{dy} \sin y \frac{dy}{dx}=1$$\quad$
- $\Rightarrow\frac{dy}{dx}[1 + \cos y] = 1$

- $\Rightarrow\frac{dy}{dx} = \frac{1}{1 + \cos y}$ , provided $\cos y \neq-1$                 
- Note that by doing have, we need not get $\frac{dy}{dx}$ as a function of $x$ only.                 
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<footer style = "font-size: 18px">$\frac{d}{d x}\left(\sec ^{-1} x\right)$ $\rightarrow$ $\frac{d}{d x}\left(\sec ^{-1} x\right)$ $\rightarrow$ <span style = "color:blue"> Implicit Differentiation </span> $\rightarrow$ Exponential Function $\rightarrow$ Graph of exp $(x)$ </footer>

<h3 id="success-stylefont-size25px-derivatives-l-4-success"><Success style="font-size:25px"> Derivatives L-4 </Success></h3>
<h3 id="primary-stylefont-size24px-exponential-function-primary"><primary style="font-size:24px"> Exponential Function </primary></h3>
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<li>$\exp (x)  =1+\frac{x}{1 !}+\frac{x^2}{2 !}+\cdots+\frac{x^n}{n !}+\cdots $</li>
<li>$ =\sum_{n=0}^{\infty} \frac{x^n}{n !} =\lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{x^k}{k !}$</li>
<li><ins>Fact </ins>: The above series converges (to a finite real number) for every $x\in R.$ Thus exp$(x)$ is defined for every real no $x$.</li>
<li><strong>Further, exp$(x)$ satisfies the following properties :</strong></li>
<li>(1) exp (0) = 1 $\quad$ (2) exp $(x)$ is an increasing function of $x$ i.e., $x, < x_2\Rightarrow exp (x_1) < exp (x_2)$.</li>
<li>(3) Domain = R  $\quad$ (4) Range of exponential fn $=R^+= (0,\infty)$ (exp$(x)$ is never zero or negative)</li>
<li>(5) lim exp$(x) = +\infty$ $\quad$ (6) lim exp$(x) = 0 x\rightarrow\infty$</li>
<li>(7) $\exp (x+y)=\exp (x) \exp (y)$</li>
<li>(Compare with $a^{m+n}=a^m \cdot a^n$ if $m, n \in \mathbb{N}$ ) Using (7) and (1), we get</li>
<li>$\quad \exp (-x)  =\frac{1}{\exp (x)}$</li>
<li>$(\because 1=\exp (0)  =\exp (x+(-x))=\exp (x) \exp (-x))\quad $ So, (6) follows from (5)</li>
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<footer style = "font-size: 18px">$\frac{d}{d x}\left(\sec ^{-1} x\right)$ $\rightarrow$ Implicit Differentiation $\rightarrow$ <span style = "color:blue"> Exponential Function </span> $\rightarrow$ Graph of exp $(x)$ $\rightarrow$ $2 < e < 3$ </footer>

### <Success style="font-size:25px"> Derivatives L-4 </Success>
### <primary style="font-size:24px"> Graph of exp $(x)$ </primary>
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- $y=exp(x)$

- Putting $x=1$, we get

- $\exp (1)  =1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots+\frac{1}{n !}+\cdots $

- $ =\sum_{k=0}^{\infty} \frac{1}{k !}$

- We know that $\exp (1)$ is 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">Implicit Differentiation $\rightarrow$ Exponential Function $\rightarrow$ <span style = "color:blue"> Graph of exp $(x)$ </span> $\rightarrow$ $2 < e < 3$ $\quad \rightarrow \quad$ $2 < e < 3$ </footer>

<h3 id="success-stylefont-size25px-derivatives-l-4-success-1"><Success style="font-size:25px"> Derivatives L-4 </Success></h3>
<h3 id="primary-stylefont-size24px--2--e-3-primary"><primary style="font-size:24px"> $ 2 < e <3$ </primary></h3>
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<p>$e=1+\frac{1}{1 !}+\frac{1}{2 !}+\cdots>1+\frac{1}{1 !}=2$ .</p>
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<p>Also, $e=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots$</p>
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<p>$<1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots$</p>
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<p>(because $\frac{1}{(n+1) !}<\frac{1}{2^n}$ for $n \geqslant 2$ )</p>
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<p>Recall: Geometric series</p>
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<p>$a+a r+a r^2+\cdots=\frac{a}{1-r} \text { if }|r|<1$</p>
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<p>Putting $a=1, r = \frac{1}{2}$ gives</p>
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<p>$1+\frac{1}{2}+\frac{1}{2^2}+\cdots=\frac{1}{1-1 / 2}=2$</p>
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<p>$\therefore \quad e<1+2=3$</p>
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<footer style = "font-size: 18px">Exponential Function $\rightarrow$ Graph of exp $(x)$ $\rightarrow$ <span style = "color:blue"> $2 < e < 3$ </span> $\quad \rightarrow \quad $ $2 < e < 3$ $\rightarrow$ Derivative of $e^x= \ exp (x)$ </footer>

<h3 id="success-stylefont-size25px-derivatives-l-4-success-2"><Success style="font-size:25px"> Derivatives L-4 </Success></h3>
<h3 id="primary-stylefont-size24px--2--e--3-primary"><primary style="font-size:24px"> $ 2 < e < 3$ </primary></h3>
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<p>Fact: It can be proved that $e$ is an irrational number.</p>
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<p>Also,</p>
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<p>$ \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^n=e .$</p>
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<p>$ \lim_{x \rightarrow 0}(1+x)^{1 / x}=e .$</p>
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<p>Let’s calculate $\lim _{h \rightarrow 0} \frac{\exp (h)-1}{h}$</p>
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<p>Notation: We write $\exp (x)$ as $e^x$.</p>
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<p>$ e^h=\exp (h)=1+\frac{h}{1 !}+\frac{h^2}{2 !}+\cdots+\frac{h^h}{n !}+\cdots$</p>
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<p>$ \Rightarrow \quad e^h-1=h+\frac{h^2}{2 !}+\frac{h^3}{3 !}+\cdots+\frac{h^h}{n !}+\cdots$</p>
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<p>$ \Rightarrow \quad \frac{e^h-1}{h}=1+\frac{h}{2 !}+\frac{h^2}{3 !}+\cdots+\frac{h^{h-1}}{n !}+\cdots \text { for } h \neq 0 . $</p>
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<p>$ \lim _{h \rightarrow 0} \frac{e^h-1}{h}=1$</p>
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<footer style = "font-size: 18px">Graph of exp $(x)$ $\rightarrow$ $2 < e < 3$ $\quad \rightarrow \quad $ <span style = "color:blue"> $2 < e < 3$ </span> $\rightarrow$ Derivative of $e^x=\exp (x)$ $\quad \rightarrow$ Example  </footer>

### <Success style="font-size:25px"> Derivatives L-4 </Success>
### <primary style="font-size:24px"> Derivative of $e^x=\exp (x)$ </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">$2 < e < 3$ $\quad \rightarrow \quad $ $2 < e < 3$ $\rightarrow$ <span style = "color:blue"> Derivative of $e^x=\exp (x)$ </span> $\rightarrow$ Example  $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Derivatives L-4 </Success>
### <primary style="font-size:24px"> Example  </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) $\quad \frac{d}{d x}\left(e^{x^2}\right)=e^{x^2} \cdot \frac{d}{d x}\left(x^2\right)=2 x e^{x^2}$.

- (3) $\quad \frac{d}{d x} \tan ^{-1}\left(e^x\right)=\frac{1}{1+\left(e^x\right)^2} \cdot \frac{d}{d x}\left(e^{x}\right)=\frac{e^x}{1+e^{2 x}}$
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<footer style = "font-size: 18px">$2 < e < 3$ $\rightarrow$ Derivative of $e^x=\exp (x)$ $\rightarrow$ <span style = "color:blue"> Example  </span> $\rightarrow$ Thank you $\rightarrow$  </footer>