Find equation of a line tangent to a curve with a given point Find equation of a line tangent to the curve at the given point:
determine an equation of a line tangent to the curve at the given point: $9(x^2 + y^2)^2 = 100xy^2$; at the point $(1, 3)$
How would I go about solving this. I know it requires implicit differentiation.
I have been having a lot of trouble differentiating and getting the derivatives.
I asked this question the Math chat room too and a few users started me off with the following: $9(x^2 + f(x)^2) = 100xf(x)^2$
I know that $f(x)^2$ would be f(x)f(x)' + f(x)'f(x). So $9(x^2 + f(x)f(x)' + f(x)'f(x)) = 100xf(x)f(x)' + f(x)'f(x)$. Past this point, I am completely confused.
 A: We start with 
$$9(x^2 + y^2)^2 = 100xy^2,$$
and we are interested in finding a tangent line. Since we are in a calculus class, we'll do this with derivatives. So let's take a derivative!
In particular, let us think of $y$ as a function of $x$ (if we wanted, we could write $y(x)$ instead of $y$ to emphasize that we're thinking of $y$ as a function of $x$). So when we differentiate $y(x)$, we just get $y'(x)$.
With this, we will apply our knowledge of the power rule (that $\frac{d}{dx} x^n = nx^{n-1}$), the chain rule (that $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$), and the product rule (that $\frac{d}{dx} f(x)g(x) = f'(x)g(x) + f(x)g'(x)$).
And by "we", I currently mean "you":
You need to try to compute
$$\frac{d}{dx} \left[ 9(x^2 + y^2(x))^2\right] = \frac{d}{dx}\left[100xy^2(x)\right],$$
(which we are currently working on with you in chat and will continue to work with you until you get through this) -- at which point I'll edit this answer.
A: Hint: $$9(x^2 + y^2)^2 = 100xy^2\\
9(2(x^2 + y^2))(2x+2y\dfrac{dy}{dx}) = 100(y^2+2xy\dfrac{dy}{dx})\\
\implies \dfrac{9x(x^2+y^2) - 50y^2}{(100x-18)y}=\dfrac{dy}{dx}$$
By implicit differentiation.
A: Using implicit differentation we get
$$18(x^2+y^2)(2x+2y\frac{dy}{dx})=100y^2+200xy \frac{dy}{dx}$$
let $x=1$, $y=3$ to get
$$18(1+9)(2+6\frac{dy}{dx})=900+600 \frac{dy}{dx}$$
solving (with not guarantee I didnt make a mistake)
$$\frac{dy}{dx}=\frac{9}{8}$$
so the slope of the line is $\frac{9}{8}$ and it passes through $(1,3)$ thus
$$y-3=\frac{9}{8}(x-1)$$
