How do you find the first and second derivative of #(x^3-3x^2+8x+18)/x#?

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Answer 1 · 2018-06-12T05:44:33

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#y=(x^3-3x^2 + 8x+18 )/x #

# y= x^3/x - (3x^2)/x + (8x)/x + 18/x #

#y = x^2 -3x + 8 +18x^(-1) #

#=> d/dx((x^3-3x^2 + 8x+18 )/x) = d/dx (x^2 -3x + 8 +18x^(-1) )#

#color(red)(d/dx ( x^n ) = nx^(n-1) #

#=> (dy)/(dx) = 2x - 3 -18x^(-2) #

#=> (d^2y)/(dx^2) = d/dx ( (dy)/(dx) ) #

#=> (d^2y)/(dx^2) = d/dx ( 2x - 3 -18x^(-2) )#

#=> (d^2y)/(dx^2) = 2 + 36x^(-3) #

Answer 2 · 2018-06-12T05:47:29

# (i) : f'(x)=2x-3-18/x^2=(2x^3-3x-18)/x^2#.

# (ii) : f''(x)=2+36/x^3={2(x^3+18)}/x^3#.

Prerequisite : #(x^n)'=nx^(n-1)#.

Let #f(x)=(x^3-3x^2+8x+18)/x#.

#:. f(x)=x^3/x-(3x^2)/x+(8x)/x+18/x, i.e., #

# f(x)=x^2-3x+8+18/x#.

#:. f'(x)=(x^2)'-3(x)'+0+18(x^-1)'#,

#=2x^(2-1)-3(1x^(1-1))+18(-1x^(-1-1))#.

# rArr f'(x)=2x-3-18/x^2=(2x^3-3x-18)/x^2#.

Similarly, #f''(x)=(f'(x))'#,

#=(2x-3-18x^-2)'#,

#=2-0-18(-2x^(-2-1))#.

# rArr f''(x)=2+36/x^3={2(x^3+18)}/x^3#.