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Differentiate/Implicit/Tangent

Heres the q:

A curve has the equation xy3+2x2y=3xy^3 + 2x^2y = 3

a Differentiate by rearranging the equation to the form y=...y = ...

b Differentiate the original equation implicitly and then find the equation of the tangent to this curve at the point (1,1)


I dont have answers in the book so im not sure if im right

for (a) I got 34×17−3x2 \frac {3}{4} \times \frac {1}{\sqrt{7-3x^2}}. Is that right?

for (b) I got to y3+x3y2dydx+4xy+2x2dydx=0y^3 + x3y^2\frac {dy}{dx} + 4xy + 2x^2\frac {dy}{dx} = 0. Is that right/what do i do next? I know the equation of the tangent, but not sure how to get the gradient from that or the other x and y value
Reply 1
Avatar for Jooeee
Jooeee
2
Factorise dy/dx out.
Reply 2
Avatar for G O D I V A
G O D I V A
OP
11
what does that do? I dont see what to do next
Reply 3
Avatar for Jooeee
Jooeee
2
Well you can rearrange for dy/dx on one side, hence you have gradient function, and you know the x and y co-ords at the point you want to find gradient for (1,1).. Sub them in and you'll get gradient.
Reply 4
Avatar for G O D I V A
G O D I V A
OP
11
you mean dydx=−y3−4xyx3y2+2x2\frac {dy}{dx} = \frac {-y^3 - 4xy}{x3y^2 +2x^2}

for (1,1) it equals -5/3

but how do i get y1 and x1?

for: 1−y1=−53(1−x1)1 - y_1 = \frac {-5}{3}(1 - x_1)
Reply 5
Avatar for Jooeee
Jooeee
2
Erm well i just use y=mx+c, and i would get,

1=-5/3+c
c=1--5/3
c=8/3

Don't know if that's any help, but i've never use the y2-y1=m(x2-x1) so don't know, sorry.
Reply 6
Avatar for G O D I V A
G O D I V A
OP
11
what do i need c for/how does that help me get the equation of the tangent?

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