How do you calculator the derivative for #(2x-5)/(x^(2)-4)#?

Answer 1

The quotient rule says that for some #h(x) = f(x)/(g(x))#, the derivative is:

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

so we get:

#h'(x) = [(x^2-4)(2) - (2x-5)(2x)]/(x^2-4)^2 = (2x^2- 8 - 4x^2 + 10x)/(x^2-4)^2# #= (-2x^2+10x-8)/(x^2-4)^2# #= [(-2x + 2)(x - 4)]/(x^2-4)^2 = [2(-x+1)(x - 4)]/(x^2-4)^2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Paisley Johnson

To find the derivative of the function ( f(x) = \frac{2x - 5}{x^2 - 4} ), you can use the quotient rule, which states that if ( f(x) = \frac{g(x)}{h(x)} ), then ( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} ). Applying the quotient rule to the given function:

[ f'(x) = \frac{(2)(x^2 - 4) - (2x - 5)(2x)}{(x^2 - 4)^2} ]

[ f'(x) = \frac{2x^2 - 8 - 4x^2 + 10x}{(x^2 - 4)^2} ]

[ f'(x) = \frac{-2x^2 + 10x - 8}{(x^2 - 4)^2} ]

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.