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Mathematics 16 Online
OpenStudy (anonymous):

if y^4=4x^3-3x+3, find dy/dx

OpenStudy (anonymous):

\[4y^3y'=12x^2-3\] solve for y'

OpenStudy (amistre64):

4y^3 y'= 12x^2 - 3 y'= (12x^2 - 3)/4y^3

OpenStudy (amistre64):

\[\sqrt{OHMM!!}\]

OpenStudy (anonymous):

or if you want more work you can always start with \[f(x)=\sqrt[4]{4x^3-3x+3}\] and do a bunch more work

OpenStudy (anonymous):

Id like as little work as possible :) im already confused enough as it is. ha

OpenStudy (anonymous):

\[\color{pink}{\text{be here now!}}\]

OpenStudy (anonymous):

So what about this one? if \[5x^2y+3xy^2+2y^3=4\], find dy/dx.

OpenStudy (anonymous):

implicit diff. pretend \[y=f(x),y'=f'(x)\] and use the product rule where applicable

OpenStudy (anonymous):

okay. so can you help me work it out step by step?

OpenStudy (anonymous):

\[5x^2y'+10xy+3y^2+3x\times 2yy'+6y^2y'=0\] then a bunch of algebra to isolate y'

OpenStudy (anonymous):

for example in the term \[5x^2y\] you need the product rule because it is a product, like if you were finding the derivative of \[5x^2\sin(x)\]

OpenStudy (anonymous):

if it were \[5x^2\sin(x)\] you would write \[10x\sin(x)+5x^2\cos(x)\] and if it were just \[5x^2f(x)\] you would write \[10xf(x)+5x^2f'(x)\] and it is the same with \[5x^2y\] you get \[10xy+5x^2y'\]

OpenStudy (anonymous):

Understood.

OpenStudy (anonymous):

good, because that is all it is, except for the annoying algebra of solving for y'

OpenStudy (anonymous):

Which i just did.

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