find dy/dx if x³+y³-3xy²-3y=15..
Answers
Answer:
\begin{gathered}x^3+y^3-3xy^2-3y=15\\\\\text{Apply derivatives on both sides}\\\\\dfrac{d}{dx}(x^3+y^3-3xy^2-3y)=\dfrac{d}{dx}(15)\\\\\\\dfrac{d}{dx}(x^3)+\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(3xy^2)-\dfrac{d}{dx}(3y)=0\\\\\\3x^2+3y^2.\dfrac{dy}{dx}-3\bigg[x.2y\dfrac{dy}{dx}+y^2\bigg]-3.\dfrac{dy}{dx}=0\\\\\\3x^2+3y^2.\dfrac{dy}{dx}-6xy\dfrac{dy}{dx}-3y^2-3.\dfrac{dy}{dx}=0\\\\\\3y^2.\dfrac{dy}{dx}-6xy\dfrac{dy}{dx}-3.\dfrac{dy}{dx}=3y^2-3x^2\end{gathered}
x
3
+y
3
−3xy
2
−3y=15
Apply derivatives on both sides
dx
d
(x
3
+y
3
−3xy
2
−3y)=
dx
d
(15)
dx
d
(x
3
)+
dx
d
(y
3
)−
dx
d
(3xy
2
)−
dx
d
(3y)=0
3x
2
+3y
2
.
dx
dy
−3[x.2y
dx
dy
+y
2
]−3.
dx
dy
=0
3x
2
+3y
2
.
dx
dy
−6xy
dx
dy
−3y
2
−3.
dx
dy
=0
3y
2
.
dx
dy
−6xy
dx
dy
−3.
dx
dy
=3y
2
−3x
2
\begin{gathered}\dfrac{dy}{dx}=\dfrac{3(y^2-x^2)}{(3y^2-6xy-3)}=\dfrac{3(y^2-x^2)}{3(y^2-2xy-1)}\\\\\\\boxed{\mathbf{\dfrac{dy}{dx}=\dfrac{y^2-x^2}{y^2-2xy-1}}}\end{gathered}
dx
dy
=
(3y
2
−6xy−3)
3(y
2
−x
2
)
=
3(y
2
−2xy−1)
3(y
2
−x
2
)
dx
dy
=
y
2
−2xy−1
y
2
−x
2