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Mathematics 22 Online
OpenStudy (anonymous):

why is dx/dy of (2x-3x^2)^5 = 12x^5 (1-3x) (2-3x)^5

OpenStudy (anonymous):

this is what i got (6) (-6x+2) (2x-3x^2)^5 =12(1-3x) (2x-3x^2)^5

OpenStudy (anonymous):

you mean \(\dfrac{dx}{dy}\) ? and \(y = (2x-3x^2)^5\), right?

OpenStudy (anonymous):

right

OpenStudy (anonymous):

Remember \(\dfrac{dx}{dy}\neq \dfrac{dy}{dx}\)

OpenStudy (anonymous):

so that, if you have y w.r.t x, then dx/dy = 1/(dy/dx)

OpenStudy (anonymous):

sorry i meant dy/dx

OpenStudy (anonymous):

thanks for noticing

OpenStudy (anonymous):

ok. hihihi

OpenStudy (imstuck):

Use the chain rule. Deal with the exponent of 5 first, then what's inside the parenthesis next.

OpenStudy (anonymous):

could u please explain it to me?

OpenStudy (imstuck):

I think OOPS left.

OpenStudy (imstuck):

Oh sure, I would be honored.

OpenStudy (anonymous):

what am i supposed to do after this: = (6) (-6x+2) (2x-3x^2)^5 =12(1-3x) (2x-3x^2)^5

OpenStudy (imstuck):

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