A bit lost on this implicit differentiation and finding the slope of a curve

[sinclairharry]

Question is to find the slope of the curve at given point :

4 x y ^(1/3) + 2 y = 56 ; (5,8)

I thought the answer was

y' = - ( y^(1/3)) / (x + 1/2)

and then sub in point of (5,8)

so the slope would equal -4 / 11

but i seem to be wrong can anyone help me?

 

[sinclairharry]

.

but why does the 56 become y ?

 

[Deleted member 4993]

but why does the 56 become y ?

Ooops .... It doesn't (if it did - then the problem would have been lot simpler!!!)

4 *x * y ^(1/3) + 2 y = 56


4 * [y^(1/3) + x * (1/3) * y^(-2/3) * y'] + 2 * y' = 0


2 * [y^(1/3) ] = - y' [1 + 2 * x * (1/3) * y^(-2/3)]


y' = -2 * [y^(1/3) ]/[1 + 2 * x * (1/3) * y^(-2/3)]


y' = -6 * y / [3 * y^(2/3) + 2 * x]


y' = -6y / (3y^2/3 + 2x) .....Now continue...

 

[sinclairharry]

thank you !

I see where i went wrong !