# Differentiating This Function

Somebody want to help with this derivative:
y = (7.75x3 + 2250) / 750x

I differentiate it with the quotient rule and get:
dy/dx = 31x / 1500 - 3 / x2

But that's wrong. It's got a zero around 5 - 6 and the original function has its minimum at (4.03, 1.12). Not sure what I did wrong.

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SammyS
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Homework Helper
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Somebody want to help with this derivative:
y = (7.75x3 + 2250) / 750x
I differentiate it with the quotient rule and get:
dy/dx = 31x / 1500 - 3 / x2

But that's wrong. It's got a zero around 5 - 6 and the original function has its minimum at (4.03, 1.12). Not sure what I did wrong.
You have two variables on the right hand side. Is that intended? If so, how is r related to x ?

MOD note: Changed the copied text to reflect changes in the OP.

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I'm sorry, I keep typing it wrong. There's only meant to be a single variable on the right-hand side. Its fixed now.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Somebody want to help with this derivative:
y = (7.75x3 + 2250) / 750x

I differentiate it with the quotient rule and get:
dy/dx = 31x / 1500 - 3 / x2

But that's wrong. It's got a zero around 5 - 6 and the original function has its minimum at (4.03, 1.12). Not sure what I did wrong.
$\displaystyle \frac{7.75x^3+2250}{750x}=\frac{7.75x^2}{750}+ \frac{2250}{750x}= \frac{31}{3000}x^2+ \frac{2250}{750x}$

The 7.75 gets divided by the 750 .

Somebody want to help with this derivative:
y = (7.75x3 + 2250) / 750x

I differentiate it with the quotient rule and get:
dy/dx = 31x / 1500 - 3 / x2

But that's wrong. It's got a zero around 5 - 6 and the original function has its minimum at (4.03, 1.12). Not sure what I did wrong.
You appear to have differentiated correctly and the derivative does indeed have a zero at x=5.256 (to 3dp).

Where did you get the (apparently false) information that the original function has a local minimum at x=4.03?

In fact, if you plug x=4.03 into the original function, you get y=0.91 (approx). Even allowing for rounding errors, the point (4.03, 1.12) isn't even close to being a point on the graph.

You appear to have differentiated correctly and the derivative does indeed have a zero at x=5.256 (to 3dp).

Where did you get the (apparently false) information that the original function has a local minimum at x=4.03?

In fact, if you plug x=4.03 into the original function, you get y=0.91 (approx). Even allowing for rounding errors, the point (4.03, 1.12) isn't even close to being a point on the graph.
Do you know what I did? I simplified the original function to get the one I posted here, just checked and they're not the same. That's why the derivative didn't match up. :surprised I think I've about got it sorted out now.