mathematics-class-12-unit-06-chapter-01-derivatives-l-1-12-its2qe7ys5k

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Relation between continuity and differentiability </primary>
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- Consider the function $f(x)=|x|, x \in \mathbb{R}$                 
- $ f(x)={\begin{cases}x,  x \geqslant 0 \\\ -x,  x<0\end{cases}}$            
- Is $f(x)$ continuous  at $x=0$ ?            
- $ \lim _{h \rightarrow 0+} f(0+h)=\lim _{h \rightarrow 0^{+}}|h|=\lim _{h \rightarrow 0^{+}} h=0 $,      
- $ \lim _{h \rightarrow 0^{-}} f(0+h)=\lim _{h \rightarrow 0^{-}}|h|=\lim _{h \rightarrow 0^{-}}(-h)=0$

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<footer style = "font-size: 18px"> $\rightarrow$ Derivatives lecture - 1 $\rightarrow$ <span style = "color:blue"> Relation between continuity and differentiability </span> $\rightarrow$ Relation between continuity and differentiability $\rightarrow$ Relation between continuity and differentiability </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Relation between continuity and differentiability </primary>
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- $\therefore \lim _{x \rightarrow 0} f(x)=0$

- Also, $f(0)=|0|=0$   
- $ \therefore f(0)=\lim _{x \rightarrow 0} f(x)$, hence $f(x)$ is cont. at $x=0$.   
- Is $f(x)$ differentiable at $x=0$ ?   
- Does $\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$ exist?

- For h ≠ 0,  ${\frac{f(0+h)-f(0)}{h}}=\frac{f(h)-f(0)}{h}=\frac{|h|-0}{h}=\frac{|h|}{h}$    $ ={\begin{cases}1  , if  h>0 \\\ -1  , if  h<0\end{cases}}$      
- $ \therefore L.H.S. \neq R.H.S.$  of $ ~ \frac{f(0+h)-f(0)}{h}$

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<footer style = "font-size: 18px">Derivatives lecture - 1 $\rightarrow$ Relation between continuity and differentiability $\rightarrow$ <span style = "color:blue"> Relation between continuity and differentiability </span> $\rightarrow$ Relation between continuity and differentiability $\rightarrow$ Theorem </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Relation between continuity and differentiability </primary>
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- Hence, $\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$ does not exist.      
- $\therefore f(x)$ is not differentiable at $x=0$.

- **Conclusion:** $f(x)$ continuous at $x=a$    
- does not imply that $f(x)$ is differentiable at $x=a$.

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<footer style = "font-size: 18px">Relation between continuity and differentiability $\rightarrow$ Relation between continuity and differentiability $\rightarrow$ <span style = "color:blue"> Relation between continuity and differentiability </span> $\rightarrow$ Theorem $\rightarrow$ Theorem </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Theorem </primary>
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- **Theorem:** If $f(x)$ is differentiable at $x=a$ then $f(x)$ must be continuous at $x=a$.

- **Proof:**  Since $f(x)$ is differentiable at $x=a$, $f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ exists.

- To check the continuity at $x=a$, we need
- $ f(a)= \lim _{x \rightarrow a} f(x)=\lim _{h \rightarrow 0} f(a+h)$

- $(\text { Put } x=a+h, \text { ie. } h=x-a,$

- $\text { As } x \rightarrow a, h \rightarrow 0)$

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<footer style = "font-size: 18px">Relation between continuity and differentiability $\rightarrow$ Relation between continuity and differentiability $\rightarrow$ <span style = "color:blue"> Theorem </span> $\rightarrow$ Theorem $\rightarrow$ Product Rule for derivatives </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Theorem </primary>
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- $\lim _{h \rightarrow 0} f(a+h)=f(a) \Leftrightarrow \lim _{h \rightarrow 0}[f(a+h)-f(a)]=0$

- But, for $h \neq 0, f(a+h)-f(a)=h\left[\frac{f(a+h)-f(a)}{h}\right]$

- Since $\lim _{h \rightarrow 0} h=0$ and $\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=f^{\prime}(a)$
  
- $ \text{ by the product rule for limits, we have}$
- $\lim _{h \rightarrow 0}[f(a+h)-f(a)]=0 \cdot f^{\prime}(a)=0 \text { }$

- Hence $f(x)$ is cont. at $x=a$.

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<footer style = "font-size: 18px">Relation between continuity and differentiability $\rightarrow$ Theorem $\rightarrow$ <span style = "color:blue"> Theorem </span> $\rightarrow$ Product Rule for derivatives $\rightarrow$ Product Rule for derivatives </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Product Rule for derivatives </primary>
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- $\frac{d}{d x}[f(x) g(x)]=f^{\prime}(x) g(x)+f(x) g^{\prime}(x) $

-  provided $f^{\prime}(x)$ \& $g^{\prime}(x)$ exists.      
-  Pf:      
- Let $u(x)=f(x) g(x)$.

- For $h \neq 0, \quad \frac{u(x+h)-u(x)}{h}$

- $ =\frac{f(x+h) g(x+h)-f(x) g(x)}{h}$

- $ =\frac{[f(x+h) g(x+h)-f(x) g(x+h)]+[f(x) g(x+h)-f(x) g(x)]}{h}$

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<footer style = "font-size: 18px">Theorem $\rightarrow$ Theorem $\rightarrow$ <span style = "color:blue"> Product Rule for derivatives </span> $\rightarrow$ Product Rule for derivatives $\rightarrow$ Example 1 </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Product Rule for derivatives </primary>
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- $\therefore \frac{u(x+h)-u(x)}{h}  ={\left[\frac{f(x+h)-f(x)}{h}\right] g(x+h)} $

- $ +f(x)\left[\frac{g(x+h)-g(x)}{h}\right]$

- Now, $\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f^{\prime}(x)$           
- $\lim _{h \rightarrow 0} g(x+h)=g(x)$

- $[\because g$ is continuous  at $x$ since it is given to be differentiable at x]

- similarly, $\lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}=g^{\prime}(x)$

- $\therefore  u^{\prime}(x)=\lim _{h \rightarrow 0} \frac{u(x+h)-u(x)}{h}=f^{\prime}(x) g(x)+f(x) g^{\prime}(x)$

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<footer style = "font-size: 18px">Theorem $\rightarrow$ Product Rule for derivatives $\rightarrow$ <span style = "color:blue"> Product Rule for derivatives </span> $\rightarrow$ Example 1 $\rightarrow$ Example 2 </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Example 1 </primary>
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- $\frac{d}{d x}[f(x) g(x)] \neq  f^{\prime}(x) g^{\prime}(x)$     
	
- **e.g.** $f(x)=g(x)=x$

- then $  f^{\prime} (x)=1= g^{\prime} (x)$

- but $f(x) g(x)=x^2$

- So, $\frac{d}{d x}[f(x) g(x)]=\frac{d}{d x}\left(x^2\right)=2 x \neq f^{\prime}(x) g'(x)$

- **e.g.**
- $f(x)  =x^3 , f^{\prime}(x)= ? $

- $ f(x)  =\frac{d}{d x}\left(x^2\right) \cdot x+x^2 \frac{d}{d x}(x) $

- $ =2 x \cdot x+x^2 \cdot 1 $

- $ =2 x^2+x^2=3 x^2 $

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<footer style = "font-size: 18px">Product Rule for derivatives $\rightarrow$ Product Rule for derivatives $\rightarrow$ <span style = "color:blue"> Example 1 </span> $\rightarrow$ Example 2 $\rightarrow$ Example 2 </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Example 2 </primary>
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- Calculate $\frac{d}{d x}\left(x^n\right)$, where $n \in \mathbb{N}$.    
- $f(x)=x^n$

- For $h \neq 0, \frac{f(x+h)-f(x)}{h}$

- $  =\frac{(x+h)^n-x^n}{h}$

- $  =\frac{x^n+ \binom n1 x^{n-1} h+ \binom n2 x^{n-2} h^2+\cdots+h^n - x^n}{h} $

- $  =\frac{h[n x^{n-1}+ \binom n2 x^{n-2} h+\cdots+h^{n-1}]}{h}$

- $  = {n x^{n-1} }$ as h $\rightarrow$ 0

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<footer style = "font-size: 18px">Product Rule for derivatives $\rightarrow$ Example 1 $\rightarrow$ <span style = "color:blue"> Example 2 </span> $\rightarrow$ Example 2 $\rightarrow$ Example 3 </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Example 2 </primary>
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- $ \therefore \frac{d}{d x}\left[x^n\right] =n x^{n-1}$  for  n $\in N $

- $ \text { eg. } \quad \frac{d}{d x}\left(x^2\right)  =2 x^{2-1}=2 x$

- $ \frac{d}{d x}\left(x^3\right) =3 x^{3-1}=3 x^2$

- $\frac{d}{d x}\left(x^4\right)  =4 x^3 .$

- **Remark:** The above formula for derivative of $x^n$ actually holds true for any $n \in \mathbb{R}$. This will be proved later.

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

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Example 3 </primary>
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- $f(x)=\frac{1}{x}, f^{\prime}(x)=$ ?

- $\frac{f(x+h)-f(x)}{h}  =\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$

- $ =\frac{x-(x+h)}{h x(x+h)}=\frac{-h}{h x(x+h)}$     
- $ =\frac{-1}{x(x+h)}, h \neq 0 .$

- $\therefore \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\frac{-1}{x^2} \quad \forall x \neq 0 .$

- $ \frac{d}{d x}\left(\frac{1}{x}\right)=\frac{-1}{x^2}, x \neq 0$

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

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Example 3 </primary>
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- Note that this agrees with same formula for
- $\frac{d}{d x}\left(x^n\right)=n x^{n-1} \text {. }$

- For $n=-1$, this gives $\frac{d}{d x}\left(x^{-1}\right)=(-1) x^{-1-1}=-\{x}^{-2}=\frac{-1}{x^2} \text { }$

- We can use product rule to calculate $\frac{d}{d x}\left(x^{-2}\right) or \frac{d}{d x}\left(x^{-3}\right)$, etc.
- eg. 
 
- $\frac{d}{d x}\left(x^{-2}\right)=\frac{d}{d x}\left(\frac{1}{x} \cdot \frac{1}{x}\right)$   
- $=\left(\frac{-1}{x^2}\right) \cdot\frac{1}{x}+\frac{1}{x} \cdot\left(\frac{-1}{x^2}\right)$   
- $ =\frac{-2}{x^3} .$

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

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Theorem </primary>
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- Suppose $f(x)$ is differentiable at $x=a$ 
	
- and let $g(x)=\frac{1}{f(x)}$ \& $f(a) \neq 0$

- Then $g^{\prime}(a)=\frac{-f^{\prime}(a)}{(f(a))^2}$

- **Proof:**

- $ \frac{g(a+h)-g(a)}{h}=\frac{\frac{1}{f(a+h)}-\frac{1}{f(a)}}{h}$

- $ =\frac{f(a)-f(a+h)}{h f(a) f(a+h)}$

- $ {=-\underbrace{\left \[\frac{f(a+h)-f(a)}{h}\right]}_{f^{\prime}(a)} \cdot \frac{1}{f(a) {f(a+h)}}}$      
- $ \therefore g'(a) = \frac{-f'(a)}{(f(a))^2}$

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<footer style = "font-size: 18px">Example 3 $\rightarrow$ Example 3 $\rightarrow$ <span style = "color:blue"> Theorem </span> $\rightarrow$ Theorem $\rightarrow$ Example 4 </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Example 4 </primary>
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- $ (uv)'= u'v+uv' \leftarrow Product Rule$       
- $ (\frac{u}{v})'= \frac{u'v-uv'}{v^2} \leftarrow Quotient Rule$     
- e.g. Derivative of $f(x)=\sqrt{x}, x>0$.     
- $ \text { If } x>0,$      
- $ \frac{f(x+h)-f(x)}{h}  =\frac{\sqrt{x+h}-\sqrt{x}}{h}$      
- $  =\frac{\sqrt{x+h}-\sqrt{x}}{h}.\frac{(\sqrt{x+h}+\sqrt{x})}{(\sqrt{x+h}+\sqrt{x})}$     
- $  =\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$   
- $  =\frac{1}{\sqrt{x+h}+\sqrt{x}} \rightarrow \frac{1}{2\sqrt{x}}$

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<footer style = "font-size: 18px">Theorem $\rightarrow$ Theorem $\rightarrow$ <span style = "color:blue"> Example 4 </span> $\rightarrow$ Example 4 $\rightarrow$ Example 5 </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Example 5 </primary>
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- Derivative of $f(x)=\sin (x)$   
- $\frac{f(x+h)-f(x)}{h}=\frac{\sin (x+h)-\sin x}{h}$

- **Recall:**     
- $ \sin C-\sin D=2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$       
- $\sin (x+h)-\sin x=2 \cos \frac{2 x+h}{2} \sin \frac{h}{2}$

- $ \therefore \frac{f(x+h)-f(x)}{h}  =\frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \frac{h}{2}}{h}$

- $ =\cos \left(x+\frac{h}{2}\right) \cdot \frac{\sin (h/2)}{h / 2}$ $ \left(\because \lim _{h \rightarrow 0} \frac{\sin h}{h}=1\right)$

- $ \therefore \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}= cosx . 1$

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<footer style = "font-size: 18px">Example 4 $\rightarrow$ Example 4 $\rightarrow$ <span style = "color:blue"> Example 5 </span> $\rightarrow$ Example 6 $\rightarrow$ Example 6 </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Example 6 </primary>
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- $ \frac{d}{d x}[\tan x]=\frac{d}{d x}\left[\frac{\sin x}{\cos x}\right]$    
- $ \text { Using Quotient rule }$   
- $ =\frac{\frac{d}{d x}(\sin x) \cos x-\sin x \frac{d}{d x}(\cos x)}{(\cos x)^2}$      
- $  =\frac{(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos ^2 x}$      
- $  =\frac{\cos ^2 x+\sin ^2 x}{\cos ^2 x}=\frac{1}{\cos ^2 x}=\sec ^2 x$     
 - $ \therefore \frac{d}{d x}[\tan x]=\sec ^2 x$

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<footer style = "font-size: 18px">Example 5 $\rightarrow$ Example 6 $\rightarrow$ <span style = "color:blue"> Example 6 </span> $\rightarrow$ Example 6 $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Derivatives L-1 </Success>
### <primary style="font-size:24px"> Example 6 </primary>
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- $ \frac{d}{d x}[\sec x]=\frac{d}{d x}\left[\frac{1}{\cos x}\right] = \frac{-\frac{d}{dx}(\cos x)}{(cosx)^2}$       
- $ =\frac{\sin x}{\cos ^2 x}=\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}$       
- $ =\tan x \sec x$     
- $ \therefore \frac{d}{d x}[\sec x] =\sec x \tan x$        
- similarly,$ \frac{d}{d x}[cosec x]=-cosec x. \cot x$      
- $ \quad \frac{d}{d x}[\cot x]=-\operatorname{cosec}^2 x$ \{Exercise}

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<footer style = "font-size: 18px">Example 6 $\rightarrow$ Example 6 $\rightarrow$ <span style = "color:blue"> Example 6 </span> $\rightarrow$ Thank you $\rightarrow$  </footer>