Finding Tangent line for a Graph with the Natural Log 
I'm really confused on how my professor did this problem. Any in depth explanation would be awesome. Thanks for your time.
 A: $$y=x^{\frac1{x}}$$
This can be a very difficult function to differentiate using our common differentiating methods. So what your professor did is take the easy way out: Get rid of the exponent.
How do we get rid of exponents? One word - logarithms. We'll use a base-$e$ logarithm (natural logarithm) because differentiating it is simple. So we just take the natural logarithm of each side:
$$\ln y=\ln x^{\frac1{x}}$$
Well, it's a property of logarithms that the exponents come to the front. Hence the exponent of $\frac1{x}$ will be moved to be the coefficient of the term:
$$\ln y=\frac1{x}\ln x$$
Awesome. Now it's easier to take the derivative. You'll need to do a little bit of implicit differentiation. Start with taking the derivative of the left side. Recall that:
$$\frac{d}{dx}\ln u=\frac1{u}\frac{du}{dx}$$
So the derivative of $\ln y$ with respect to $x$:
$$\frac{d}{dx}\ln y=\frac1{y}\frac{dy}{dx}=\frac{y'}{y}$$
(Nicer to see: $y'=\frac{dy}{dx}$)
Now for the right side, something similar. Except we have to use the product rule. The product rule is:
$$\frac{d}{dx}uv = u \frac{dv}{dx} + v \frac{du}{dx}$$
Awesome. We have:
$$\frac{d}{dx}\frac1{x}\ln x = \frac1{x} \underbrace{\left( \frac{d}{dx}\ln x \right)}_{\frac1{x}}+\ln x \underbrace{\left( \frac{d}{dx} \frac1{x} \right)}_{-\frac1{x^2}}=\frac1{x^2}-\frac1{x^2}\ln x$$
We can now combine both sides:
$$\frac{y'(x)}{y(x)}=\frac1{x^2}-\frac1{x^2}\ln x$$
I say, don't bother even reducing. Just plug in the point to get the slope ($y'$):
$$\frac{y'(1)}{(1)}=\frac1{(1)^2}-\frac1{(1)^2}\ln (1)$$
$$y'(1)=1$$
And that's our slope! Now just plug it in to the normal slope equation:
$$y=mx+b$$
$m=1$, and we are at the point $(1, 1)$ therefore $b$ is:
$$1=(1)(1)+b$$
$$b=0$$
We are therefore left with the general equation:
$$\therefore y=x$$
Here's what they look like

Cheers!
-Shahar
