**(A) Second-Order Differentiation**

**1.**When a function

*y*=

*x*

^{3 }+

*x*

^{2 }– 3

*x*+ 6 is differentiated with respect

to

*x,*the derivative $\frac{dy}{dx}=3{x}^{2}+2x-3$

**2.**The

**second function**$\frac{dy}{dx}$ can be differentiated again with respect

to

*x*. This is called the

**second derivative**of

*y*with respect to

*x*

and can be written as $\frac{{d}^{2}y}{d{x}^{2}}$ .

**3.**Take note that $\frac{{d}^{2}y}{d{x}^{2}}\ne {\left(\frac{dy}{dx}\right)}^{2}$ .

**For example,**

If

*y*= 4*x*^{3}– 7*x*^{2}+ 5*x*– 1,**The first derivative**$\frac{dy}{dx}=12{x}^{2}-14x+5$

**The second derivative**$\frac{{d}^{2}y}{d{x}^{2}}=24x-14$

**(B) Turning Points, Maximum and Minimum Points**

**(a) At Turning Points**

*A*and*B*,**(b) At Maximum Point**

*A*,**(c) At Minimum Point**

*B*,**Example 1**

**(Maximum Value of Quadratic Function)**

*y*= 3

*x*(4 –

*x*), calculate

**(a)**the value of

*x*when

*y*is a maximum,

**(b)**the maximum value of

*y*.

*Solution:***(a)**

**(b)**

$\begin{array}{l}y=12x-3{x}^{2}\\ \text{When}x=2,\\ y=12\left(2\right)-3{\left(2\right)}^{2}\\ y=12\end{array}$

**Example 2 (Determine the Turning Points and Second Derivative Test)**

*y*= 2

*x*

^{3 }+ 3

*x*

^{2 }– 12

*x*+ 7 and determine the nature of these turning points.

*Solution:*6

*x*^{2}+ 6*x*– 12 = 0*x*

^{2}+

*x*– 2 = 0

(

*x*– 1)*(**x*+ 2) = 0*x*= 1 or

*x*= –2

When

*x*= 1*y*= 2(1)

^{3}+ 3(1)

^{2}– 12(11) + 7

*y*= 0

**(1, 0) is a turning point.**

When

*x*= –2*y*= 2(–2)

^{3}+ 3(–2)

^{2}– 12(–2) + 7

*y*= 27

**(–2, 27) is a turning point.**

**Hence, the turning point (1, 0) is a minimum point.**

$\begin{array}{l}\text{When}x=-2,\\ \frac{{d}^{2}y}{d{x}^{2}}=12\left(-2\right)+6=-180\text{(negative)}\end{array}$

**Hence, the turning point (–2, 27) is a maximum point.**