# Higher order derivatives , Definition , Formula , Solved Examples

Let y = f(x)

First order derivative

$\large \frac{dy}{dx} = f'(x) = y_1$

Second order derivative

$\large \frac{d^2y}{dx^2} = f”(x) = y_2$

Third order derivative

$\large \frac{d^3y}{dx^3} = f ”'(x) = y_3$

nth order derivative

$\large \frac{d^ny}{dx^n} = f^{n}(x) = y_n$

Illustration :

(i) If y3 − 3ax2 + x3 = 0 , show that

$\large \frac{d^2y}{dx^2} + \frac{2a^2 x^2}{y^5} = 0$

(ii) If y = (sin−1x)2 + k sin−1x , show that

$\large (1-x^2)\frac{d^2y}{dx^2} – x\frac{dy}{dx} = 2$

Solution: (i) y3 − 3ax2 + x3 = 0

$\large 3y^2 \frac{dy}{dx} – 6ax + 3 x^2 = 0$

$\large y^2 \frac{dy}{dx} – 2ax + x^2 = 0$

$\large y^2 \frac{dy}{dx} = 2ax – x^2 = 0$

$\large \frac{dy}{dx} = \frac{2ax – x^2}{y^2}$  …(ii)

Again diff. w. r. to x , we get

$\large \frac{d^2y}{dx^2} = \frac{y^2(2a – 2x)-(2ax-x^2)2y\frac{dy}{dx}}{y^4}$

$\large \frac{d^2y}{dx^2} = \frac{y^2(2a – 2x)-(2ax-x^2)2y(\frac{2ax – x^2}{y^2})}{y^4}$

$\large \frac{d^2y}{dx^2} = \frac{y^3(2a-2x)-2(2ax-x^2)^2}{y^5}$

$\large \frac{d^2y}{dx^2} = -\frac{2 a^2 x^2}{y^5}$

$\large \frac{d^2y}{dx^2} + \frac{2 a^2 x^2}{y^5} = 0$

(ii) If y = (sin−1x)2 + k sin−1x

$\large \frac{dy}{dx} = 2 sin^{-1}x . \frac{1}{\sqrt{1-x^2}} + k \frac{1}{\sqrt{1-x^2}}$

$\large \sqrt{1-x^2}\frac{dy}{dx} = 2 sin^{-1}x + k$

Again, differentiating both sides w.r.t. x , we get

$\large \sqrt{1-x^2}\frac{d^2y}{dx^2} + \frac{1}{2\sqrt{1-x^2}}(-2x) \frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} + 0$

$\large (1-x^2)\frac{d^2y}{dx^2} – x \frac{dy}{dx} = 2$

Exercise :

(i) Find the derivative w. r. to x of the following functions

(a) $\large y = log(sin\sqrt{x^2 + 1})$

(b) $\large y = \frac{logx}{1+xlogx}$

(c) y = xx2

(d) y = cos−1√cosx

(e) $\large y = log(\frac{1+\sqrt{x}}{1-\sqrt{x}})$

(f) x cos y = sin (x + y)

(g) $\large y = \frac{5 tan^2 x – 1}{tanx}$

(ii) If y = x logy , prove that

$\large \frac{dy}{dx} = \frac{y^2}{x(y-x)}$

(iii) If $\large x = \frac{3at}{1+t^2}$ and , $\large y = \frac{3at^2}{1+t^3}$

Show that $\large \frac{dy}{dx} = \frac{(1+t^2)^2}{(1+t^3)^2} . \frac{t(2-t^3)}{(1-2t^2)}$