what is the first and second derivative of X^2/x-3?

Question

what is the first and second derivative of X^2/x-3

Answer 1

Use the quotient rule:

f(x) = g(x)/h(x) ---------> f'(x) = (g'h - h'g)/h^2

In this problem, g(x) = x^2, h(x) = x-3
Therefore, g'(x) = 2x, h'(x) = 1

Substituting these values, you get:

f'(x) = [(2x(x-3) - x^2]/(x-3)^2

Simplifying, you get, f'(x) = (x^2-6x)/(x-3)^2

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To solve for f''(x), take the derivative of f'(x). Use the quotient rule again:

g(x) = x^2-6x ----------> g'(x) = 2x-6
h(x) = (x-3)^2 ----------> h'(x) = 2(x-3)(1) = 2x-6

Plugging into the equation, you get:

f''(x) = [(2x-6)(x-3)^2 - (2x-6)(x^2-6)]/(x-3)^2

Simplifying, you get:

f''(x) = (2x-6)[(x-3)^2 - (x^2 - 6)]/(x-3)^2 = 2(x-3)/(x-3)^2[(x-3)^2 - (x^2 - 6)]

Therefore, f''(x) = 2[(x^2-6x+9-x^2+6)]/(x-3) = (30-12x)/(x-3)

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Hope this helps

Answer 2

f(x) = x^2/x-3 = x^2(x-3)^-1
f'(x) = -x^2(x-3)^-2 +2x(x-3)(x-3)^-2
f'(x) = (-x^2 +2x(x-3))(x-3)^-2
f'(x) = (-x^2 +2x^2-3x)(x-3)^-2
f'(x) = (x^2-3x)(x-3)^-2
f'(x) = x(x - 3)(x-3)^-2
f'(x) = x(x-3)^-1

f'(x) = x/(x-3)

f''(x) = -x(x-3)^-2 + (x-3)(x-3)^-2
f''(x) = (-x + x-3)(x-3)^-2

f''(x) = -3/(x-3)^2