dy/dx for (y^2)x = 4, etc.

[mattflint50]

did I do this right

I am finding dy/dx

(y^2)x=4
(y^2)x-4=0
radical (x-4)=y

t=x-4
y=(t)^(.5)
dy/dt=.5t^(-.5)
dt/dx=1
dy/dx=1/2(x-4)^(-.5)

thanks

 

[skeeter]

no ... you did not.

do you know how to find the derivative using implicit differentiation?

y²x = 4

y² = 4/x

2y(dy/dx) = -4/x²

dy/dx = -4/(2x²y) = -2/(x²y)

 

[mattflint50]

ok, could you show me how to do this one please.

d^2y/dx^2 of (x^2+y^2=16)

I think it would help me to understand
thank you very much

 

[mattflint50]

for an answer I got - (2x)^2/y

 

[galactus]

The first implicit derivative of \(\displaystyle x^{2}+y^{2}=16\) is

\(\displaystyle \L\\\frac{-x}{y}\)

Now, do it again using quotient rule:

\(\displaystyle \L\\\frac{y(-1)-x\frac{dy}{dx}}{y^{2}}\)

But \(\displaystyle \frac{dy}{dx}=\frac{-x}{y}\)

\(\displaystyle \L\\\frac{x^{2}}{y^{3}}-\frac{1}{y}\)