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

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

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 $f(x)$

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

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

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

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

Suppose $f(x)$ in an increasing function in an interval $I=(a, b)$.

ie. if $x_1, x_2 \in I, x_1 < x_2$ the $f(x) \leq f\left(x_2\right)$

What can we say about $f^{\prime}(x)$ if $f(x)$ is a differentiable $f x$ ?

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

Right hand derivative $=\lim _{h \rightarrow 0+} \frac{f(x+h)-f(x)}{h} \geqslant 0$

Here $f(x+h) \geqslant f(x)$ since $f$ is increasing.

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

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

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

$<0 \text { on }(0,\infty)$

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 $x_0$.

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,

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

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

$f\left(\frac{3}{2}\right)=-\frac{1}{4} \quad \therefore f(x)$ 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 $[a, b]$.

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)=$\begin{cases}x, 0 \leq x<\frac{1}{2} \\ 0, x=\frac{1}{2} \\ 1-x, \frac{1}{2}<x \leqslant 1\end{cases}.$

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.

(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 $f^{\prime}(c)=0$.

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)

But $f^{\prime}(x)=1$ for all $x \in(1,4)$ Thus there is no $c \in(1,4)$ for which

$f^{\prime}(c)=0 \text {. }$

However, this example does not contradict the Rolle's theorem because $f(x)$ is not cont. on the closed interval $[1,4]$.