Question

Assume that y is a function of x. Find y' = dy/dx for

Answer 1

\[x^{3}+y^{3} = 4 \]

Answer 2

same as before, chain-rule "y", then solve for dy/dx :)

Answer 3

\(\bf x^3+y^3=4\implies 3x^2+3y^2\cfrac{dy}{dx}=0\implies \cfrac{dy}{dx}=\cfrac{\cancel{-3} x^2}{\cancel{3} y^2}
\)

Answer 4

\[\frac{ d }{ dx }(x)^{3}+y^{3} = \frac{ d }{ dx }(4)\]

\[\frac{ d }{ dx }x^{3} = 3x^{2}\]

\[\frac{ d }{ dx }(4) = 0 \]

\[\frac{ d }{ dx }[y]^{3}*\frac{ d }{ dx }*y = 3y^{2}*\frac{ dy }{ dx }\]

\[3x^{2}+3y^{2}*\frac{ dy }{ dx } = 0 \]

\[\frac{ -3x^{2} }{ 3y^{2} } = \frac{ dy }{ dx } = \frac{ -x }{ y }\]

Answer 5

\[\frac{ -x^{2} }{ y^{2} } = \frac{ dy }{ dx }\]

Answer 6

What does this necessarily mean though? this is giving us the change in x and y

Answer 7

@jdoe0001

Answer 8

@KyanTheDoodle

Answer 9

yes, a derivative is the slope equation.. so, that's what we get :)

Answer 10

so this shows how x and y are changing with respect to each-other?

Answer 11

well, a slope is a rise/run of a line, yes, in this case, the tangent line at some point
and a derivative gives you just that

Answer 12

so it's like just the slope of the tangent line then okay

Answer 13

a derivative, is just the function to get the slope at some given point, so
if you were to need the slope of that equation at some point when x = 3 and y = 7
then that'd be \(\bf \cfrac{-3^2}{7^2}\)

Answer 14

okay thanks

Answer 15

yw