Some basic rules of differentiation are as follows, where #u# and #v# are functions of #x#:
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Addition / subtraction rule
If #y=u+-v#, #dy/dx=d/dx(u)+-d/dx(v)#
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Chain rule
#dy/dx=dy/(du)xx(du)/dx#
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Product rule
If #y=uv#, #dy/dx=u(dv)/dx+v(du)/dx#
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Quotient rule
If #y=u/v#, #dy/dx=(v(du)/dx-u(dv)/dx)/v^2#
Let's get started,
#f(x)=(x^2 - 8)/(x^2+3) #
First, differentiate #f(x)#,
#f'(x)=([d/dx(x^2-8)] (x^2+3)-[d/dx(x^2+3)] (x^2-8))/(x^2+3)^2# ( Quotient rule )
#color(white)(f'(x))=((2x) (x^2+3)-(2x) (x^2-8))/(x^2+3)^2#
#color(white)(f'(x))=((2x) (cancel(x^2)+3cancel(-x^2)+8))/(x^2+3)^2#
#color(white)(f'(x))=((2x)(11))/(x^2+3)^2#
#color(white)(f'(x))=(22x)/(x^2+3)^2#
Since, it is an stationary point, #d/dx[f(x)]=0#,
#(22x)/(x^2+3)^2=0#
#color(white)(xx/xxx.)22x=0#
#color(white)(xxxxxx)x=0#
Find the #y#-coordinate of the stationary point by substituting #x=0# into #f(x)#,
#y= ((0)^2 - 8)/((0)^2+3)#
#y=-8/3#
Hence, the co-ordinate of the stationary point is #(0,-8/3)#.
To find the nature of the stationary point, differentiate #f'(x)#,
#f''(x)=([d/dx(22x)] (x^2+3)^2-[d/dx(x^2+3)^2] (22x))/(x^2+3)^4# ( Quotient rule )
#color(white)(f''(x))=((22)(x^2+3)^2-(4x)(x^2+3)(22x))/(x^2+3)^4# ( Product rule )
#color(white)(f''(x))=((22)(x^2+3)[(x^2+3)-(4x)(x)])/(x^2+3)^4#
#color(white)(f''(x))=((22)(3-3x))/(x^2+3)^3#
#color(white)(f''(x))=(66-66x^2)/(x^2+3)^3#
Substitute #x=0# into #f''(x)#,
#f''(x)=(66-66(0)^2)/((0)^2+3)^3#
#color(white)(f''(x))=66/27 > 0# ( minimum point )
Hence, the stationary point #(0,-8/3)# is, in fact, a minimum point and not an inflexion point.
Check:
graph{(x^2 - 8)/(x^2+3) [-10, 10, -5, 5]}
