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)=(3x^2 + 8x + 5)/(4-x)#
First, differentiate #f(x)#,
#f'(x)=([d/dx(3x^2 + 8x + 5)] (4-x)-[d/dx(4-x)](3x^2 + 8x + 5))/(4-x)^2#
#color(white)(f'(x))=((6x+8) (4-x)-(-1)(3x^2 + 8x + 5))/(4-x)^2#
#color(white)(f'(x))=(-6x^2+16x+32+3x^2 + 8x + 5)/(4-x)^2#
#color(white)(f'(x))=-(3x^2-24x-37)/(x-4)^2#
Since, it is an stationary point, #d/dx[f(x)]=0#,
#-(3x^2-24x-37)/(x-4)^2=0#
#-3x^2+24x+37=0#
#color(white)(xxxxxx/xxxxxx)x=-1.32291 or 9.32291#
Find the #y#-coordinate of the stationary points by substituting #x=-1.32291 or 9.32291# into #f(x)#,
#y=-0.063 or -63.937#
Hence, the co-ordinates of the stationary points are #(-1.323,-0.063)# and #(9.323,-63.937)#.
To find the nature of the stationary points, differentiate #f'(x)#,
#f''(x)=-([d/dx (3x^2-24x-37)] (x-4)^2-[d/dx(x-4)^2] (3x^2-24x-37))/(x-4)^4#
#color(white)(f''(x))=-((6x-24) (x-4)^2-(2)(x-4)(1)(3x^2-24x-37))/(x-4)^4#
#color(white)(f''(x))=-((x-4)[(6x-24)(x-4)-(2)(1)(3x^2-24x-37)])/(x-4)^4#
#color(white)(f''(x))=(6x^2-48x+96-6x^2+48x+74)/(x-4)^3#
#color(white)(f''(x))=-170/(x-4)^3#
Let #x=-1.32291#,
#f''(x)=-170/(1.32291-4)^3#
#color(white)(f''(x))=8.86054>0# ( minimum )
Let #x=9.32291#,
#f''(x)=-170/(9.32291-4)^3#
#color(white)(f''(x))=-1.12720<0# ( maximum )
Hence, minimum#(-1.323,-0.063)# and maximum#(9.323,-63.937)#.
Check:
graph{(3x^2 + 8x + 5)/(4-x) [-320, 320, -160, 160]}
