How do you find the derivative of #f(x) = (x^2-4)/(x-1)#?

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Answer 1 · 2016-06-06T18:35:57

In a function #f(x) = (g(x))/(h(x))#, the derivative is given by #((g'(x) xx h(x)) - (g(x) xx h'(x)))/(h(x))^2#

We must determine the derivatives of the numerator and of the denominator,

Numerator:

Let #g(x) = x^2 - 4#

By the power rule, #g'(x) = 2x#.

Denominator:

Let #h(x) = x - 1#

By the power rule, #h'(x) = 1#
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We can now differentiate with the quotient rule.

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

#f'(x) = ((2x(x - 1)) - ((x^2 - 4)1))/(x - 1)^2#

#f'(x) = (2x^2 - 2x - x^2 + 4)/(x - 1)^2#

#f'(x) = (x^2 - 2x + 4)/(x^2 - 2x + 1)#

Thus, the derivative of #f(x) = (x^2 - 4)/( x - 1)# is #dy/dx = (x^2 - 2x + 4)/(x^2 - 2x + 1)#

Hopefully this helps!