C4 Differentiation helps Watch

piccolo12345
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#1
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#1
A curve is defined by the equation  4x^2 + y^2 = 4 + 3xy
Find the gradient at the point (1,3) on this curve.

Can someone please go through this question for me step by step?
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*MJ*
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Do you know that d/dx f(y) = f'(y)dy/dx?

Apply this principle - use the product rule for the differentiation of 3xy.

Make dy/dx the subject.

Sub in your values of x and y.

Obtain gradient.
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piccolo12345
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(Original post by *MJ*)
Do you know that d/dx f(y) = f'(y)dy/dx?

Apply this principle - use the product rule for the differentiation of 3xy.

Make dy/dx the subject.

Sub in your values of x and y.

Obtain gradient.
I've got
8x + 2y\frac{dy}{dx} = 3x \frac{dy}{dx} + 3y

but i'm not sure how to make dy/dx the subject with two of them?
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piccolo12345
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ahh na got it combining using brackets.. thanks
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Farhan.Hanif93
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(Original post by piccolo12345)
ahh na got it combining using brackets.. thanks
Just to clarify, it's called factorising and you've been doing it since GCSE
/Pedant
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Kameo
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 4x^2 + y^2 = 4 + 3xy



8x + 2y\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}




2y\frac{dy}{dx} - 3x\frac{dy}{dx} = 3y -8x



\frac{dy}{dx} (2y -3x) = 3y -8x




\frac{dy}{dx} = \frac{3y - 8x}{2y-3x}



\frac{dy}{dx} = \frac{1}{3}
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