Can't get this implicit differentiation I've been working at this implicit differentiation problem for a little over an hour now, and I, nor my friends can figure it out. The question reads "Find the equation of the tangent line to the curve (a lemniscate) $2(x^2+y^2)^2=25(x^2−y^2)$ at the point (3,1). Write the equation of the tangent line in the form $y=mx+b.$"
Every time that we do it we get a ridiculous number for the slope (${150}/{362}$)
 A: We have
$2(x^2+y^2)^2=25(x^2−y^2)$.
Since
$(x^2)' = 2 x \  dx$,
I would differentiate this as
$2(2(x^2+y^2)(x^2+y^2)')
=25(2x\ dx - 2y\ dy)
$
or
$4(x^2+y^2)(2x\ dx-2 y\ dy)
=25(2x\ dx - 2y\ dy)
$
or
$8(x(x^2+y^2)dx-y(x^2+y^2)dy)
=50x\ dx-50y\ dy
$
or
$(8x(x^2+y^2)-50x)dx
=(8y(x^2+y^2)-50y)dy
$
or
$\dfrac{dy}{dx}
=\dfrac{8x(x^2+y^2)-50x}{8y(x^2+y^2)-50y}
$.
Putting
$x=3$
and $y=1$,
$\begin{array}\\
\dfrac{dy}{dx}
&=\dfrac{8x(x^2+y^2)-50x}{8y(x^2+y^2)-50y}\\
&=\dfrac{24(9+1)-150}{8(8+1)-50}\\
&=\dfrac{90}{22}\\
&=\dfrac{45}{11}\\
\end{array}
$.
If the line is
$y = mx+b$,
$m = 45/11$
and
$b = y-mx
=1-3m
=1-135/11
=-124/11
$.
All errors are (obviously)
my fault,
and I will accept
all appropriate
punishments.
A: Familiarize yourself with the notion of a gradient if you haven't done so already.
Here are the equations for the gradient so that you can check your solutions when you are ready:

Plug in the appropriate coordinates to get the gradient.
A: Let's consider $y$ to be a function of $x$ (usually written "$y=y(x)$") for $x$ close to $3$. Then
$$2(x^2+y(x)^2)^2=25(x^2-y(x)^2)$$
so by differentiating on $x$ (using basically the chain rule), we obtain
$$4(x^2+y(x)^2)(2x+2y(x)y'(x))=25(2x-2y(x)y'(x))$$
Substitute $x=3$ and $y(x)=y(3)=1$, and you will find the value of $y'(3)=-9/13$ (at least that's what I found). This means that the slope of the tangent is $-9/13$, so the equation for the tangent is
$$y=y(3)+y'(3)(x-1)=1-(9/13)(x-1)=-\frac{9}{13}x+22/13$$
