# Math Help - implicit differentiation help

1. ## implicit differentiation help

The exact wording of the problem is: "Determine the slope of the normal line to the curve $x^{3}+xy^{2} = 10y$ at the point $(2,1)$"

Now since finding the slope is the same thing as finding the derivative, I assumed that I would need to take the derivative of the problem, aka use implicit differentiation.

I got: $3x^{2}+y^{2}y{}'+x2yy{}' = 10y{}'$
Which would mean that $\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3x^{2}}{-y^{2}-x2y+10}$

However, when I plugged in $x$ and $y$, I got $\frac{12}{5}$, which was not among the answer choices of $0$, $2$, $-\frac{7}{3}$, $-\frac{6}{13}$, or $\frac{1}{2}$.

2. Originally Posted by TheMathTham
The exact wording of the problem is: "Determine the slope of the normal line to the curve $x^{3}+xy^{2} = 10y$ at the point $(2,1)$"

Now since finding the slope is the same thing as finding the derivative, I assumed that I would need to take the derivative of the problem, aka use implicit differentiation.

I got: $3x^{2}+ {\color{red}y^{2}y{}'} +x2yy{}' = 10y{}'$ Mr F says: The red term is wrong.

Which would mean that $\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3x^{2}}{-y^{2}-x2y+10}$

However, when I plugged in $x$ and $y$, I got $\frac{12}{5}$, which was not among the answer choices of $0$, $2$, $-\frac{7}{3}$, $-\frac{6}{13}$, or $\frac{1}{2}$.
..

3. Originally Posted by TheMathTham
The exact wording of the problem is: "Determine the slope of the normal line to the curve $x^{3}+xy^{2} = 10y$ at the point $(2,1)$"

Now since finding the slope is the same thing as finding the derivative, I assumed that I would need to take the derivative of the problem, aka use implicit differentiation.

I got: $3x^{2}+y^{2}y{}'+x2yy{}' = 10y{}'$
Which would mean that $\frac{\mathrm{dy} }{\mathrm{d} x}=\frac{3x^{2}}{-y^{2}-x2y+10}$

However, when I plugged in $x$ and $y$, I got $\frac{12}{5}$, which was not among the answer choices of $0$, $2$, $-\frac{7}{3}$, $-\frac{6}{13}$, or $\frac{1}{2}$.
The slope of the normal to the tangent is given by the equation km = -1.

So if the slope of the tangent is k= 13/6 then the slope of the tangent is m = -6/13.

$\frac{d(xy^2)}{dx}= y^2 + 2xy\frac{dy}{dx}$

4. Originally Posted by xalk

$\frac{d(xy^2)}{dx}= y^2 + 2xy\frac{dy}{dx}$
Can you walk me through the steps on how you got that? I'm confused

5. Originally Posted by TheMathTham
Can you walk me through the steps on how you got that? I'm confused
$\frac{d(xy^2)}{dx}=\frac{d}{dx}\left(xy^2\right)$.
Apply the product rule.

6. Originally Posted by TheMathTham
Can you walk me through the steps on how you got that? I'm confused
$\frac{d(xy^2)}{dx}=\frac{dx}{dx}y^2 + x\frac{d(y^2)}{dx} = 1y^2 +2xy\frac{dy}{dx}$

7. Originally Posted by xalk
$\frac{d(xy^2)}{dx}=\frac{dx}{dx}y^2 + x\frac{d(y^2)}{dx} = 1y^2 +2xy\frac{dy}{dx}$
I think I must have learned implicit differentiation incorrectly, because I'm still lost. Why was it left as $\frac{d(xy^2)}{dx}=1y^2 +2xy\frac{dy}{dx}$ ? I don't recall ever learning how to use $\frac{d(xy^2)}{dx}$ in implicit differentiation.

The derivative of $x^{3}$ is $3x^{2}$.
The derivative of $xy^{2}$ is $(y^{2} + 2xy)\frac{\mathrm{d} y}{\mathrm{d} x}$
The derivative of $10y$ is $10\frac{\mathrm{d} y}{\mathrm{d} x}$

I'm assuming that the one highlighted in blue is where I'm messing up. $u=x, u{}'=1, v=y^{2}, v{}'=2y$
$vu{}'+uv{}'=y^{2}+2xy$

EDIT:
Never mind, I realized that dy/dx isn't included in the y^2 portion of the product rule

8. Originally Posted by TheMathTham
I think I must have learned implicit differentiation incorrectly, because I'm still lost. Why was it left as $\frac{d(xy^2)}{dx}=1y^2 +2xy\frac{dy}{dx}$ ? I don't recall ever learning how to use $\frac{d(xy^2)}{dx}$ in implicit differentiation.

The derivative of $x^{3}$ is $3x^{2}$.
The derivative of $xy^{2}$ is $(y^{2} + 2xy)\frac{\mathrm{d} y}{\mathrm{d} x}$
The derivative of $10y$ is $10\frac{\mathrm{d} y}{\mathrm{d} x}$

I'm assuming that the one highlighted in blue is where I'm messing up. $u=x, u{}'=1, v=y^{2}, v{}'=2y$
$vu{}'+uv{}'=y^{2}+2xy$

EDIT:
Never mind, I realized that dy/dx isn't included in the y^2 portion of the product rule
$v{}' = 2yy{}'$ and not $v{}' =2y$

because $v{}' = \frac{dv}{dx}$ and according to chain rule

$\frac{dv}{dx} = \frac{dv}{dy}\frac{dy}{dx}$

9. Originally Posted by xalk
$v{}' = 2yy{}'$ and not $v{}' =2y$

because $v{}' = \frac{dv}{dx}$ and according to chain rule

$\frac{dv}{dx} = \frac{dv}{dy}\frac{dy}{dx}$
I was referring to ${\color{red} y^{2}} + 2xy\frac{\mathrm{d} y}{\mathrm{d} x}$ actually, but thanks anyways