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
Which class you are presently in?
Choose an option. You can change your option at any time.
You will be solving questions and growing your critical thinking skills.   