Differential equation (product/quotient)

[recoil33]

I have several questions, so i'll post these as i go.

1.

y = (5x+2)²/((x+5)³)²

u = (5x+2)²
u' = 7(5x+3)⁶ . (5)
v = (x+5)³
v' = (3(x+5)²).(1)

Quotient = (u(x)*v'(x) - v(x)* u'(x) / v(x)²)

(5x+2)⁷ . 3(x+5)² - (x+5)³ . 7(5x+5)⁶ .(5) / ((x+5)³)²

[ If i knew how to put this under i would ]

What have i done wrong here? not exactly sure ;s

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2.

If i have an equation with 2 brackets on the numerator, and only 1 on the denominator will i use the product rule for the numerator then the quotient? Or, will i expand the brackets then use the quotient rule?

Example:

(4x+9)(6x+1) / (8x+3)

u = (4x+9)(6x+1)
u' = (4x+9)(6) + (6x+1)(4)
v = (8x+3)
v' = (8)

Am i doing the right thing here? Any advice will help thank you.
[If my question is not clear please tell me]

Thanks in advance, recoil33

 

[JThompson]

u = (5x+2)²
u' = 7(5x+3)⁶ . (5)

I'm assuming there's a typo-- u=(5x+2)^{7}, right?

Quotient = (u(x)*v'(x) - v(x)* u'(x) / v(x)²)

The quotient rule is (u'(x)*v(x)-u(x)*v'(x))\over (v(x))^{2}; you reversed the order of differentiation. You take the derivative of the function in the numerator first.

If i have an equation with 2 brackets on the numerator, and only 1 on the denominator will i use the product rule for the numerator then the quotient? Or, will i expand the brackets then use the quotient rule?

You can do it either way. The work you posted is correct.

 

[HallsofIvy]

But what do they have to do with differential equations?

 

[recoil33]

@ JThompson

Yes, the equation was sapose to be

u = (5x +2)⁷
u' = 7(5x+2)⁶

The equation will be incorrect if i accidently put u'(x)*v(x) and v'(x)*u(x) in the wrong order?

@Hallsofivy

Sorry, bit new to the terminology of the mathematics.

I guess it's not a differential equation?, although was it in the right section? (calc and beyond)


Thanks JThomas for your help and thanks for your input ivy ;)

 

[JThompson]

The equation will be incorrect if i accidently put u'(x)*v(x) and v'(x)*u(x) in the wrong order?

Yes. The derivative will be of the opposite sign.
v'(x)*u(x)-u'(x)*v(x)= -(u'(x)*v(x)-v'(x)u(x))

Good luck.