How do you differentiate #y=(3x)/(x^2-3)#?

Answer 1

Christopher Allers

The derivative is #dy/dx=(-3(3+x^2))/(x^2-3)^2#

This thequotient of two functions

The derivative is #(u/v)'=(u'v-uv')/v^2#

Here. #u=3x#, #u'=3#

and #v=x^2-3#, #v'=2x#

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

#=(3x^2-9-6x^2)/(x^2-3)^2#

#=(-9-3x^2)/(x^2-3)^2#

#=(-3(3+x^2))/(x^2-3)^2#

Sign up to view the whole answer

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

Sign up with email

Answer 2

Axel Alvarez

To differentiate the function y=(3x)/(x^2-3), you can use the quotient rule. The quotient rule states that if you have a function in the form of u/v, where u and v are functions of x, then the derivative is (v * du/dx - u * dv/dx) / v^2. Applying this rule to the given function, we get:

dy/dx = [(x^2 - 3) * d(3x)/dx - 3x * d(x^2 - 3)/dx] / (x^2 - 3)^2

Now, we differentiate each term:

d(3x)/dx = 3 d(x^2 - 3)/dx = 2x

Substitute these into the quotient rule formula:

dy/dx = [(x^2 - 3) * 3 - 3x * 2x] / (x^2 - 3)^2

Simplify:

dy/dx = (3x^2 - 9 - 6x^2) / (x^2 - 3)^2

dy/dx = (-3x^2 - 9) / (x^2 - 3)^2

Sign up to view the whole answer

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

Sign up with email

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.