So with implicit differentiation, you are going to be taking the derivative of every variable, in the entire equation. Every time you take the derivative of a variable, you have it's rate of change multiplied on the right. In this case, dx or dy. 

The result of the derivative is:

The first step is to create the  term that you are looking to solve for. This is done by dividing the entire equation by  to get it on the bottom of the fraction. After distributing this division to each term, the dx in the first term will cancel with itself, and you will be left with one term that is multiplied by . At that point, you want to get the term with  onto its own side. This can be accomplished by subtracting the  to the right side of the equation. 

The result so far:

Then to finish getting  on its own, you divide  to the right side, ending up with:

So now looking at the question, we know that , so in order to figure out  we need to plug  into the equation.

This gives us .

So this gives us two possible answers:

So the derivative of a natural log  is always equal to  or one over whatever is inside the natural log. In this case  is inside the natural log, so the derivative of  should be:

But since the inside of the natural log is a function as well, this is the chain rule and the derivative of the natural log will be multiplied by the derivative of the inside, in this case , which is . 

So the final derivative is 

This problem involves using product rule, and chain rule.

By product rule,

Then, using power rule, and ten chain.

Another application of chain rule to get at the angle,

Taking the derivative of the angle and then simplifying, we get

The derivative of the function is equal to

and was found using the following rules:

, , , , 

Note that the chain rule is used for the exponential function (the secant is the inner function) and for the cosine function (the linear term is the inner function). 

The derivative of the function is equal to

and was found using the following rules:

, , , 

Note that the chain rule was used on the natural logarithm derivative and the derivative of the cosine function.

The first step in finding our second derivative is finding the first.

For our function  we note that it is a composite function, and therefore requires the use of the chain rule. The chain rule states if  then .

In our case this becomes:

To find the second derivative we must take the derivative of this function we've now computed. To do this we will require both the chain rule as stated, and the product rule since our function is of the form , our derivative will be of the form .

Therefore, our second derivative is:

or equivalently,

We start by taking the derivative of both sides of the equation, and proceeding as follows,

.

First step is to figure out what  is. This is done by inputting  wherever there is an  in , which gives you 

.

Then you start on the outermost function, the sine function, and work your way inwards.

Again, chain rule for  is:

,

which gives us for this problem

So implicit differentiation means that you take the derivative of every variable in the same way, with its respective rate of change, dx or dy in this case, coming out multiplied on the end.

You take the derivative of the whole equation and the result is:

The first step to solve for dy/dx is to create a dy/dx term, which means that you need to divide the entire equation by dx, to get it on the bottom. Each term becomes divided by dx, so the dx cancels with itself and you are left with two terms that have . The result is:

At this point, you want to get  on its own side, so this involves subtracting the two terms that aren't multiplied by  onto the right side of the equation.

After that, you factor out the  and end up with:

Then, the last step is to divide over the two term block to the other side to get  on its own.

The chain rule of the derivative always deals with the composition of two or more functions.

In this case we can identify two, 

and 

.

So  is the composition of these two such that: 

With chain rule, you always start on the outermost function and work your way inward, which in this case is: 

Always the derivative of the outermost evaluated at the inner, multiplied by the derivative of the inner.