Question

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

Answer 1

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

Answer 2

4y^3 y'= 12x^2 - 3

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

Answer 3

\[\sqrt{OHMM!!}\]

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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
\[5x^2\sin(x)\]

Answer 12

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
\[10xy+5x^2y'\]

Answer 13

Understood.

Answer 14

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

Answer 15

Which i just did.