# Math Help - differentiation

1. ## differentiation

Find the equation using the tangent line to the curve defined by the equation at the specific point.

(x^2+y^2)^2= 4x^2y; (-1,1)

Use implicit differentiation to verify that the equation implicitly defines a solution of the differential equation.

y^2= 1+Ce^-2x; y(y'+y)=1

My teacher just assigned theses and did not explain what to do im lost please help. I believe derivatives are used as well as the chain rule...i think

2. Implicit differentiation can be very tricky the first time through.

I'll try to explain it the best that I can with your first example

So we have:
$(x^2+y^2)^2= 4x^{2y}$
Simplify it a little bit:
$(x^2+y^2)= 4x^y$
Now we take derivatives of both sides
$\frac{d}{dx}(x^2+y^2)= \frac{d}{dx}(4x^y)$
Now here's the tricky part. Any time we take the derivative of a multivariable function like this one we consider y to be a function of x. So for the sake of understanding how implicit differentiation works, I'll replace every occurence of y with y(x). However, this isn't what you should be doing in your homework.
$\frac{d}{dx}(x^2+y(x)^2)= \frac{d}{dx}(4x^{y(x)})$
Now we use properties of the derivative to make our job easier:
$\frac{d}{dx}(x^2)+\frac{d}{dx}(y(x))^2= \frac{d}{dx}(4x^{y(x)})$
Notice the first term contains only x, so we can take our usual derivative:
$2x+\frac{d}{dx}(y(x))^2= \frac{d}{dx}(4x^{y(x)})$
However, the other terms contain y(x), so we have to use chain rule:
Remember: $\frac{d}{dx}(y(x)) = \frac{dy(x)}{dx} * \frac{d}{dy(x)}$
So whenever we see a y(x) (or rather a y), we take the derivative with respect to why and multiply it by $\frac{dy(x)}{dx}$
So using the above argument and chain rule
$2x+\frac{dy(x)}{dx}*(y(x)^2)'= 4(\frac{dy(x)}{dx}*y'(x)*x^{y(x)}\log{x} + \frac{x^{y(x)}}{x})$
Now we replace y(x) with y and take derivatives:
$2x+\frac{dy}{dx}*2y= 4*(\frac{dy}{dx}*1*x^{y}\log{x} + \frac{x^{y}}{x})$
And now you just solve algebraically for $\frac{dy}{dx}$