Mathematics
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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