mathematics-class-12-unit-06-chapter-05-derivatives-l-5-12-hx0kg2btuda

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Example-1 </primary>
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- Eqn. of the parabola $y^2=4 a x$ can be written in the parametric form as

- $x=a t^2, y=2 a t$

- So, $\frac{d x}{d t}=2 a t, \quad \frac{d y}{d t}=2 a$

- $\therefore \frac{d y}{d x}=\frac{2 a}{2 a t}=\frac{1}{t} \text {. }$

- directly, $\quad \frac{d}{d x}\left(y^2\right)=\frac{d}{d x}(4 a x) \Rightarrow 2 y \frac{d y}{d x}=4 a$

- $\Rightarrow \frac{d y}{d x}=\frac{4 a}{2 y}=\frac{2 a}{y}$

- Putting $y=2 a t, \frac{d y}{d x}=\frac{2 a}{2 a t}=\frac{1}{t}$  .

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<footer style = "font-size: 18px"> $\rightarrow$ Derivatives lecture-5 $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-2 $\rightarrow$ Example-3 </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Example-2 </primary>
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- eg. Find $\frac{d y}{d x}$ if

- $x=a(\cos \theta+\theta \sin \theta)$

- $y=a(\sin \theta-\theta \cos \theta)$

- $\frac{d y}{d x}=\frac{\frac{d y}{d\theta}}{\frac{d y}{d\theta}}  =\frac{a(\cos \theta-\cos \theta+\theta \sin \theta)}{a(-\sin \theta+\sin \theta+\theta \cos \theta)}$   
- $=\tan \theta .$

- **Higher order derivatives:**

- suppose $y=f(x)$ is differentiable and the derivative $f^{\prime}(x)$ is a differentiable function. Then the derivative of $f^{\prime}(x)$ is called the and derivative of $f(x)$
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<footer style = "font-size: 18px">Derivatives lecture-5 $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Example-3 $\rightarrow$ Sign of derivatives </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Example-3 </primary>
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- and in denoted by $f^{\prime \prime}(x)$ or $\frac{d^2 y}{d x^2}$ ,

- similarly we can define higher order derivative ( 3rd, 4th derivatives) also.

- eg. Let $y=A \cos x+B \sin x,$  A and B are constant

- show that $\frac{d^2 y}{d x^2}+y=0$.

- $ \frac{d y}{d x}=-A \sin x+B \cos x$

- $ \frac{d^2 y}{d x^2}=-A \cos x-B \sin x=-y$

- $\therefore \frac{d^2 y}{d x^2}+y=0 .$

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

<h3 id="success-stylefont-size25px-derivatives-l-5-success"><Success style="font-size:25px"> Derivatives L-5 </Success></h3>
<h3 id="primary-stylefont-size24px-sign-of-derivatives-primary"><primary style="font-size:24px"> Sign of derivatives </primary></h3>
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<p>Suppose $f(x)$ in an increasing function in an interval $I=(a, b)$.</p>
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<p>ie. if $x_1, x_2 \in I, x_1 < x_2$ the $f(x) \leq f\left(x_2\right)$</p>
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<p>What can we say about $f^{\prime}(x)$ if $f(x)$ is a differentiable $f x$ ?</p>
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<p>$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$</p>
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<p>Right hand derivative $=\lim _{h \rightarrow 0+} \frac{f(x+h)-f(x)}{h} \geqslant 0$</p>
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<p>Here $f(x+h) \geqslant f(x)$ since $f$ is increasing.</p>
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<footer style = "font-size: 18px">Example-2 $\rightarrow$ Example-3 $\rightarrow$ <span style = "color:blue"> Sign of derivatives </span> $\rightarrow$ Sign of derivatives $\rightarrow$ Example-3 </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Sign of derivatives </primary>
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- $\text { Left-hand derivative }=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

- Here both numerator & denominator are negative.

- $\therefore \text { L.H.D } \geqslant 0$

- Thus, if $f(x)$ is a differentiable function and is increasing on $I=(a, b)$, the

- $f^{\prime}(x) \geqslant 0 \text {. }$

- Similarly, for a decreasing for which  di ff'bb, $f^{\prime}(x) \leqslant 0$.
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<footer style = "font-size: 18px">Example-3 $\rightarrow$ Sign of derivatives $\rightarrow$ <span style = "color:blue"> Sign of derivatives </span> $\rightarrow$ Example-3 $\rightarrow$ Local minima and maxima </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Example-3 </primary>
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- eg. $f(x)=x^2$ 
- $f(x)$ is decreasing in the interval $(-\infty, 0)$ and increasing in the inverse $(0, \infty)$.

- Thus, $f^{\prime}(x) \leq 0$ on $(-\infty, 0)$

- $\geqslant 0 \text { on }(0, \infty)$

- Verification: $\quad f^{\prime}(-x)=2 x <0 \text { on }(-\infty,0)$  
- $<0 \text { on }(0,\infty)$

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<footer style = "font-size: 18px">Sign of derivatives $\rightarrow$ Sign of derivatives $\rightarrow$ <span style = "color:blue"> Example-3 </span> $\rightarrow$ Local minima and maxima $\rightarrow$ Local minima and maxima </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Local minima and maxima </primary>
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- Suppose $f(x)$ is a given function. A point $x_0$ is said to be a local minimum / maximum of $f(x)$ is there is an interval $(a, b)$ containing $x_0$ such that

- $f\left(x_0\right) \leqslant f(x) \quad \forall x \in(a, b)$

- $\left(f\left(x_0\right) \geqslant f(x) \quad \forall x \in(a, b)\right)$

- For $f(x)=x^2, \quad x_0=0$ is a local minimum $x_0$ is a local minimum if

- $f(x)$ is decreasing in an interval to the left of $x_0$, and is increasing in an interval to the right of $x_0$.

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<footer style = "font-size: 18px">Sign of derivatives $\rightarrow$ Example-3 $\rightarrow$ <span style = "color:blue"> Local minima and maxima </span> $\rightarrow$ Local minima and maxima $\rightarrow$ Example-4 </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Example-4 </primary>
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- eg:

- $\begin{aligned}& f(x)=x^2-3 x+2 \\\\& f^{\prime}(x)=2 x-3 \\\\& f^{\prime}(x)= \begin{cases}0 & \text { at } x=3 / 2 \\\\& -x<3 / 2 \\\\& -x>3 / 2\end{cases}\end{aligned} $

- $\therefore$ By the 1st derivative test $f(x)$ has a local minimum at $x=3 / 2$.

- Infarct,

- $f(x) =x^2-2 \cdot \frac{3}{2} x+\left(\frac{3}{2}\right)^2+2-\left(\frac{3}{2}\right)^2$

- $=\underbrace{\left(x-\frac{3}{2}\right)^2}_{\geqslant 0}-\frac{1}{4} \quad \therefore f(x) \geqslant-\frac{1}{4}$

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<footer style = "font-size: 18px">Local minima and maxima $\rightarrow$ Local minima and maxima $\rightarrow$ <span style = "color:blue"> Example-4 </span> $\rightarrow$ Example-4 $\rightarrow$ Remark </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Example-4 </primary>
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- $f\left(\frac{3}{2}\right)=-\frac{1}{4} \quad \therefore f(x)$ takes the minimum value at $x=3 / 2$.

- Important theorem: Any continuous function $f(x)$ on a closed interval $[a, b]$ attains its minimum as well as maximum values on $[a, b]$.

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<footer style = "font-size: 18px">Local minima and maxima $\rightarrow$ Example-4 $\rightarrow$ <span style = "color:blue"> Example-4 </span> $\rightarrow$ Remark $\rightarrow$ Condition for continuity </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Remark </primary>
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- ie. for continuous $f(x)$ on $[a, b]$,
- the exist $x_1, x_2 \in[a, b]$ s.t.
- $ f\left(x_1\right) \leqslant f(-x) \leqslant f\left(x_2\right) \text { for all } x \in[a, b] \text {. } $

- we'll not look at the proof of this then but note that the assumptions are necessary.             
- $\rightarrow ~ $ Continuity in necessary:                
- f(x)=$\begin{cases}x,  0 \leq x<\frac{1}{2} \\\ 0,  x=\frac{1}{2} \\\ 1-x,  \frac{1}{2}<x \leqslant 1\end{cases}.$

- No max. value.

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<footer style = "font-size: 18px">Example-4 $\rightarrow$ Example-4 $\rightarrow$ <span style = "color:blue"> Remark </span> $\rightarrow$ Condition for continuity $\rightarrow$ Rolle's Theorem </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Condition for continuity </primary>
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- **Closed interval is necessary**.

- The result is not true for open interval.

- e.g. consider $f(x)=\frac{1}{x}$ on $(0,1)$

- $\lim _{x \rightarrow+\infty} f(x)=+\infty \text {. }$

- $f(x)$ has no max. value on $(0,1)$ even though it is continuous.
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<footer style = "font-size: 18px">Example-4 $\rightarrow$ Remark $\rightarrow$ <span style = "color:blue"> Condition for continuity </span> $\rightarrow$ Rolle's Theorem $\rightarrow$ Condition for  Rolle's theorem </footer>

### <Success style="font-size:25px"> Derivatives L-5 </Success>
### <primary style="font-size:24px"> Rolle's Theorem </primary>
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- suppose $f(x)$ is a function define l on $[a, b]$ satisfying the following three conditions:

- (i) $f(x)$ is continuous on close interval $[a, b]$

- (ii) $f(x)$ is differentiable on open interval $(a, b)$

- (iii) $f(a)=f(b)$.

- Then there exists at last one point $c \in(a, b)$ such that $f^{\prime}(c)=0$.

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<footer style = "font-size: 18px">Remark $\rightarrow$ Condition for continuity $\rightarrow$ <span style = "color:blue"> Rolle's Theorem </span> $\rightarrow$ Condition for  Rolle's theorem $\rightarrow$ Condition for  Rolle's theorem </footer>

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### <primary style="font-size:24px"> Condition for  Rolle's theorem </primary>
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- The conditions are necessary:

- $f(x)= \begin{cases}4 & \text { if } x=1 \\\ x & \text { if } 1<x \leq 4\end{cases}$
- $f(x)$ is not cont. of $x=1$ other than $x=1$,

- f(x) is cont. everywhere else.

- Also f(x) is differentiable (1,4). And f(1) = f(4)

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<footer style = "font-size: 18px">Condition for continuity $\rightarrow$ Rolle's Theorem $\rightarrow$ <span style = "color:blue"> Condition for  Rolle's theorem </span> $\rightarrow$ Condition for  Rolle's theorem $\rightarrow$ Thank you </footer>