# The Second Derivative

Contents

## Summary

• Second derivative is when we differentiate $\frac { dy }{ dx }$.
• We write second derivative of y with respect to as: $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$
• $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad \neq \quad { (\frac { dy }{ dx } ) }^{ 2 }$
• Stationary Points are obtained by solving the equation $\frac { dy }{ dx } \quad =\quad 0$.
• Maximum point: $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad <\quad 0$
• Minimum point: $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad >\quad 0$

#### Second derivative

Differentiation is the process that we use to find the gradient of a point on the curve. We define a gradient function as $\frac { dy }{ dx }$  and call it a derivative or differential coefficient of y with respect to x. The value of $\frac { dy }{ dx }$  at a given point is the gradient of the curve and hence it is also the gradient of the tangent at that point. This applies to any point along the curve.

#### Higher Derivative

A higher Derivative which could be the second derivative or the third derivative is basically calculated when we differentiate a derivative one or more times i.e

Consider a function $y\quad =\quad { x }^{ 4 }\quad +\quad 3{ x }^{ 2 }\quad +\quad 2x\quad +5$,  differentiating with respect to x, we get: $\frac { dy }{ dx } \quad =\quad 4{ x }^{ 3 }\quad +\quad 6x\quad +\quad 2\quad \quad \quad \quad \Rightarrow \quad \quad eq\quad 1$  which is another function of x.

Now if we differentiate eq 1 further with respect to x, we get: $\frac { d }{ dx } \left[ \frac { dy }{ dx } \right] \quad =\quad 12{ x }^{ 2 }\quad +\quad 6\quad \quad \quad \quad \Rightarrow \quad \quad eq\quad 2$

This eq 2 is called second derivative of y with respect to x, and we write it as: $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$

Similarly, we can find third derivative of y: $\frac { { d }^{ 3 }y }{ d{ x }^{ 3 } }$   and so on.

Note: we can not write higher derivatives in the form: $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad \neq \quad { (\frac { dy }{ dx } ) }^{ 2 }$

As ${ (\frac { dy }{ dx } ) }^{ 2 }$  means square of the derivative not the second derivative.

Let’s move on to see where exactly do we need to work out second derivatives.

#### Stationary Points

Stationary points are points on any curve where gradient of the curve is zero. These could be maximum points (point B) or the minimum points (point C) as shown in Fig 1. Stationary Points are obtained by solving the equation $\frac { dy }{ dx } \quad =\quad 0$.

However, once we have found the stationary point, to determine the nature (maximum or minimum) of this point we have to work out the second derivative of the equation of the curve.

Suppose we have a curve with equation y = f(x) which has a maximum point at A as shown in Fig 2. Now we can clearly see that the gradient of this curve is increasing before the maximum point and is decreasing after the maximum point.

On the other hand if we are not given a graph of the curve, then to find the nature of the stationary point we take a second derivative of y which will be $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$  and put x = a, if $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad <\quad 0$, then we say that f(x) has a maximum value at a.

Similarly, if the curve with equation y = f (x) which has a minimum point at B as shown in Fig 3. We clearly see that the gradient of this curve is decreasing before the minimum point and is increasing after the minimum point.

In such a case where we are not given a graph of the curve, to find the nature of the stationary point we take a second derivative of y which will be $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$  and put x = b, if $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad >\quad 0$,  then we say that f(x) has a minimum value at b.

#### Example #1

Q. The equation of a curve is $y\quad =\quad 5{ x }^{ 4 }\quad +\quad 2{ x }^{ 3 }\quad -\quad 6$

i) Express $\frac { dy }{ dx }$  and $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$  in terms of x.

ii) Find the stationary point and determine the nature of this stationary point.

Solution:

i) $y\quad =\quad 5{ x }^{ 4 }\quad +\quad 2{ x }^{ 3 }\quad -\quad 6$ $\frac { dy }{ dx } \quad =\quad 20{ x }^{ 3 }\quad +\quad 6{ x }^{ 2 }$ $\frac { dy }{ dx } \quad =\quad 20{ x }^{ 3 }\quad +\quad 6{ x }^{ 2 }\quad \quad \quad \quad \quad \quad \quad (1)$ $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad =\quad 60x\quad +\quad 12\quad \quad \quad \quad \quad \quad \quad (2)$

ii) Equate $\frac { dy }{ dx }$  to 0 $20{ x }^{ 3 }\quad +\quad 6{ x }^{ 2 }\quad =\quad 0$ $2{ x }^{ 2 }(10x\quad +\quad 3)\quad =\quad 0$

Hence: $x\quad =\quad \frac { -3 }{ 10 }$

When $x\quad =\quad \frac { -3 }{ 10 }$ $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad =\quad 60x\quad +\quad 12$ $=\quad 60(\frac { -3 }{ 10 } )\quad +\quad 12(\frac { -3 }{ 10 } )$ $=\quad -21.6$

Since $\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad <\quad 0$, $x\quad =\quad \frac { -3 }{ 10 }$  is a maximum point