Home > Calculus > Tangent Line to a Curve

If #y = 8/(3x^2)#, what is the value of #y'#?

Answer 1

Scarlett Allison

There is no problem that I can see in the reasoning used.

If you check your answer using the quotient rule, you get the same derivative.

Here's the check:

#y' = (0 xx 3x^2 - 6x xx 8)/(3x^2)^2#

#y' = (-48x)/(9x^4)#

#y' = -16/(3x^3)#

We get the same derivative (since # -16/3x^(-3) = -16/(3x^3))#, using a different method.

Hopefully this helps!

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

Answer 2

Luke Alvarez

There are many routes to the correct answer. Your method is fine.

I would have thought of the details this way:

#8/(3x^2) = 8/3 * 1/x^2 = 8/3x^-2#.

I get the exact from you got:

#y' = -16/3x^-3#.

Another possibility (that I find needlessly complicated) is to use the product rule on #8/3*x^-2#

#y' = d/dx(8/3) * x^-2 + 8/3 * d/dx(x^-2)#

# = 0 * x^-2 + 8/3 * (-2x*-3)#

# = -16/3x^-3#.

Needlessly complicated? I think so.

Incorrect? No it is not incorrect.

(If I try to go to Chicago, there are many ways to get there. If I end up in Chicago, the way I chose worked.)

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

Answer 3

Ezekiel Alexander

The value of y' is -16x/(3x^2)^2 or -16x/(9x^4).

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.