Find the derivative y′(x) implicitly

Hi!

$\displaystyle \text{A.}\,3xy^3-4x = 10y^2$

A little help with solving this please. I'll happily accept tips on solving implicit problems, considering I am weak at them.

Thank you!

$\displaystyle 3y^3 + x\cdot\frac{d}{dx}\left(y^3\right) + y^3\frac{d}{dx}(x) -\frac{d}{dx}(4x) = 20yy\prime$

$\displaystyle 3y^3+3xy^2y\prime + y^3 -4 = 20yy\prime$

$\displaystyle 4y^3+3xy^2y\prime - 4 = 20yy\prime$

$\displaystyle y\prime(3xy^2-20y)= 4-4y^3$

$\displaystyle y\prime = \frac{4-4y^3}{3xy^2-20y}$

Re: Find the derivative y′(x) implicitly

$\displaystyle \left[3x\cdot3y^2y\prime+y^3\cdot3\right]-4 = 20yy\prime$

$\displaystyle 9xy^2y\prime +3y^3 - 4 = 20yy\prime$

$\displaystyle y\prime= \frac{4-3y^3}{9xy^2-20y}$

Re: Find the derivative y′(x) implicitly

Quote:

Originally Posted by

**Unreal** Hi!

$\displaystyle \text{A.}\,3xy^3-4x = 10y^2$

A little help with solving this please. I'll happily accept tips on solving implicit problems, considering I am weak at them.

Thank you!

$\displaystyle 3y^3 + x\cdot\frac{d}{dx}\left(y^3\right) + y^3\frac{d}{dx}(x) -\frac{d}{dx}(4x) = 20yy\prime$

the derivative, with respect to x, of $\displaystyle 3xy^3$ is $\displaystyle 3y^3+ 3x(y^3)'= 3y^3+ 9xy^2y'$. You do not have the "3" that was in the original term in the second part.

I don't know where that "$\displaystyle y^3\frac{d}{dx}(x)$ comes from. You had already done that in first "3y^3[/tex]". The derivative, with respect to x is, in fact -4 so the left side is

$\displaystyle 3y^3+ 9xy^2y'- 4$. The right is 20yy' as you say so you have $\displaystyle 3y^3+ 9xy^2y'- 4= 20yy'$

Subtract 10yy' and $\displaystyle 3y^3- 4$ from both sides to get $\displaystyle 9xy^2y'- 20yy'= (9xy^2- 20y)y'= 4- 3y^3$

Your only error appears to be that missing "3".

Quote:

$\displaystyle 3y^3+3xy^2y\prime + y^3 -4 = 20yy\prime$

$\displaystyle 4y^3+3xy^2y\prime - 4 = 20yy\prime$

$\displaystyle y\prime(3xy^2-20y)= 4-4y^3$

$\displaystyle y\prime = \frac{4-4y^3}{3xy^2-20y}$