Question

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

Answer 1

Shown below

Explanation:

`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) )`

Use the power rule:

`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

`(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`.

Explanation:

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`.