c3 product rule Watch

vinsta
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#1
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i manage to differenciate using the product rule but when i get to the end its the factorising that gets me, is there any method to factorising these derivatives??
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Kolya
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#2
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Err, look for a common factor?
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zrancis
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#3
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Can you give an example?

If you mean something like this:

y=x^2(x-1){1.5}

\frac{dy}{dx}=2x(x-1)^{1.5}+1.5x^2(x-1)^{0.5}

The trick is spotting a common factor. in this case it is x(x-1)^{0.5} So if we factor that out from our derivative:

= x(x-1)^{0.5}[2(x-1)] + x(x-1)^{0.5}[1.5x]

= x(x-1)^{0.5}[2(x-1)+1.5x]

= x(x-1)^{0.5}[3.5x-1] and you're done.
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vinsta
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#4
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(Original post by zrancis)
Can you give an example?

If you mean something like this:

y=x^2(x-1){1.5}

\frac{dy}{dx}=2x(x-1)^{1.5}+1.5x^2(x-1)^{0.5}

The trick is spotting a common factor. in this case it is x(x-1)^{0.5} So if we factor that out from our derivative:

= x(x-1)^{0.5}[2(x-1)] + x(x-1)^{0.5}[1.5x]

= x(x-1)^{0.5}[2(x-1)+1.5x]

= x(x-1)^{0.5}[3.5x-1] and you're done.
well i have 18x^2(1+3x^2)^2+2(1+3x^2)^3
and i cannot seem 2 factorise it
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Kolya
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#5
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You can take out a factor of 2(1+3x^2)^2
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vinsta
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#6
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ok then so can u show me the steps to differenciating this using the product rule x^3(2x+6)^4 ??
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Kolya
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#7
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See that x^3(2x+6)^4 = 2^4 x^3 (x+3)^4 = 16x^3(x+3)^4 . Therefore (x^3(2x+6)^4)' = (16x^3(x+3)^4)' = 16\cdot 3 \cdot x^2 (x+3)^4 + 16x^3 \cdot 4 \cdot (x+3)^3 Do you want to have a go at factorizing that?
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vinsta
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#8
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(Original post by Lusus Naturae)
See that x^3(2x+6)^4 = 2^4 x^3 (x+3)^4 = 16x^3(x+3)^4 . Therefore (x^3(2x+6)^4)' = (16x^3(x+3)^4)' = 16\cdot 3 \cdot x^2 (x+3)^4 + 16x^3 \cdot 4 \cdot (x+3)^3 Do you want to have a go at factorizing that?
ah, yes i understand it now thanks
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