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Core 3 - Quotient Rule

Need help on question 2a :confused:
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Original post by Cdp3
Need help on question 2a :confused:


What have you tried? Can you show your working?
Differentiate function both the functions and plug them into the formula and simplify, you do know the chain rule method right?
Just a refresher on the quotient rule:

if y=u(x) / v(x), then dy/dx =[ v(x) * u'(x) -u(x) * v'(x)]/ [v(x)]^2,

where u'(x) and v'(x) denote the derivatives of u(x) and v(x) wrt x respectively.

Peace.
Reply 4
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Original post by Mr M
What have you tried? Can you show your working?


Yes
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Reply 5
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Original post by Dilzo999
Differentiate function both the functions and plug them into the formula and simplify, you do know the chain rule method right?


I have to use the quotient rule but yes I do
Reply 6
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Original post by Cdp3
Yes


u=(3x−2)2⇒dudx=2(3x−2)\displaystyle u=(3x-2)^2 \Rightarrow \frac{du}{dx}=2(3x-2)

This line is incorrect. Can you see your mistake?
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Original post by notnek
u=(3x−2)2⇒dudx=2(3x−2)\displaystyle u=(3x-2)^2 \Rightarrow \frac{du}{dx}=2(3x-2)

This line is incorrect. Can you see your mistake?


No?
Reply 8
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Notnek
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Original post by Cdp3
No?

You've used the chain rule but forgot to multiply by the derivative of the stuff inside the brackets.

ddx(f(x)n)=nf(x)n−1×f′(x)\displaystyle \frac{d}{dx}\left(f(x)^n\right) = nf(x)^{n-1} \times f'(x)
(edited 9 years ago)
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Original post by notnek
u=(3x−2)2⇒dudx=2(3x−2)\displaystyle u=(3x-2)^2 \Rightarrow \frac{du}{dx}=2(3x-2)

This line is incorrect. Can you see your mistake?


Is this right now?
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Original post by Cdp3
...


Yes. Although it may be easier to differentiate via chain rule than expanding out when function is more complex.
Original post by Cdp3
I have to use the quotient rule but yes I do


You have not understood the hint

You will want to use the chain rule to differentiate the numerator

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