Ucale

# Higher order derivatives

When derivative of any function itself differentiable, with the further derivative . Any derivative beyond the first derivative can be referred to as a higher order derivative.

If y = f(x) then $\displaystyle \frac { dy }{ dx }$ , the derivative of y with respect to x, is itself, in general , a function of x and can be differentiated again. To fix up the idea, we shall call  $\displaystyle \frac { dy }{ dx }$ w.r.t. x as the first order derivative of y with respect to x and

the derivative of $\displaystyle \frac { dy }{ dx }$ w.r.t. as the second order derivative of y w.r.t. x and will be denoted by $\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ .

Similarly the derivative of $\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$ w.r.t. x will be termed as the third order derivative of y w.r.t. x and will be denoted by $\displaystyle \frac { { d }^{ 3 }y }{ d{ x }^{ 3 } }$

and so on. the nth order derivative of y w.r.t. x will be denoted by $\displaystyle \frac { { d }^{ n }y }{ d{ x }^{ n } }$

If y=f(x), then the other alternative notations for

$\displaystyle \begin{matrix} \frac { dy }{ dx } & \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } & \frac { { d }^{ 3 }y }{ d{ x }^{ n } } \\ { y }_{ 1 } & { y }_{ 2 } & { y }_{ 3 } \\ y' & y'' & y''' \end{matrix}\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \begin{matrix} { D }_{ y } & { { D }^{ 2 } }_{ y } & { { D }^{ 3 } }_{ y } \\ f'\left( x \right) & f''\left( x \right) & f'''\left( x \right) \end{matrix}$

The values of these derivatives at x=a are denoted by

$\displaystyle { y }_{ n }\left( a \right) ,{ y }^{ n }\left( a \right) ,{ D }^{ n }y\left( a \right) ,{ y }^{ n }\left( a \right) \quad or\quad { \left( \frac { { d }^{ n }y }{ d{ x }^{ n } } \right) }_{ x=a }$

### Example

$\displaystyle If\quad y=\sin ^{ -1 }{ x } ,\quad show\quad that\quad \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\frac { x }{ { \left( 1-{ x }^{ 2 } \right) }^{ 3/2 } }$

Solution:

We have $\displaystyle y=\sin ^{ -1 }{ x }$ . On differentiating w.r.t. x , we get

$\displaystyle \frac { dy }{ dx } =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } }$

Again differentiating w.r.t. x we get

$\displaystyle \frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } =\frac { d }{ dx } \left( \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } \right) \\ =\frac { d }{ dx } \left[ { \left( 1-{ x }^{ 2 } \right) }^{ -1/2 } \right] \\ =-\frac { 1 }{ 2 } { \left( 1-{ x }^{ 2 } \right) }^{ -3/2 }.\frac { d }{ dx } { \left( 1-{ x }^{ 2 } \right) }^{ -3/2 }\\ =-\frac { 1 }{ 2{ \left( 1-{ x }^{ 2 } \right) }^{ 3/2 } } \left( -2x \right) \\ =\frac { x }{ 2{ \left( 1-{ x }^{ 2 } \right) }^{ 3/2 } }$

April 18, 2019
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