How to do Implicit Differentiation (Calculus)
For the First Derivative

Question

darkknight · Answer

First of all, implicit Differentiation is a way of how to find the derivative (dy/dx) when the equation is not equal to y.
What I mean by this is that if we have an equation such as y=2x+4, we can use the constant rule to find the derivative of this, which is 2. In more complex functions, that are set equal to y, such as 
$$y=2x^2(5x^3)+22x$$
 We can use other rules such as product and quotient rules of derivatives, but these rules are not enough if the function is not equal to y.

darkknight · Answer

To find the derivatives for functions that not solved for y, such as 2y^2+x = 22xy
We have to do something called implicit differentiation.

Now how does this work?

Basically what we do is we follow a series of steps, these steps are:

1) find derivatives on both sides of the equation.
2) When finding derivatives of y, we have to multiply by dy/dx or y' (these are the same thing, and they mean the derivative of y).

darkknight · Answer

So let's get into some examples, 
Let's take into consideration the equation of 
$$y^2+3x^2-4=5y^5$$
Let's take the derivatives for both sides
So the derivative of y^2 is 2y which we know by the power rule, but remember that step 2 is that we have to multiply by dy/dx when finding derivatives of y.
So then the derivative is 
$$2y (dy/dx)$$ for y^2
using power rule again, the derivative of $$3x^2 = 6x$$ 
Also using constant rule, the derivative of a constant number such as -4 is 0
So on the left side we have $$2y (dy/dx) + 6x+0$$
Now on the right side of the equation, we have 
$$5y^5$$ Using power rule we know that the derivative of this is $$25y^4$$But that is not all, because we still have to do step 2, which is when finding derivatives of y, we need to multiply by dy/dx
So then the derivative of the right side is actually $$25y^4(dy/dx)$$
We can now put all these into an equation
We have  $$2y (dy/dx) + 6x+0 = 25y^4(dy/dx)$$

darkknight · Answer

But we are not done yet, because we need to solve for dy/dx
So what we do is that we take all the terms that have dy/dx in them and put them on one side of the equation, so what I am going to do is subtract 
$$2y(dy/dx)$$ on both sides so that all dy/dx terms are on the right side.
Now our equation for the derivative looks like this
$$6x=25y^4(dy/dx)-2y(dy/dx)$$
Lets factor out dy/dx on the right, so now we have
$$6x = (25y^4-2y)(dy/dx)$$
Now we solve for dy/dx
$$dy/dx = 6x/(25y^4-2y)$$ We can simplify this further but this is what dy/dx equals in this case

darkknight · Answer

Okay, next example. Lets take the equation 
$$2x^3+4y^3 = 12xy$$

First we find derivatives of each individual side. 
So using power rule, the derivative of 2x^3 is 6x^2. And the the derivative of 
$$4y^3 = 12y^2$$ but remember we have to multiply by dy/dx every time we are finding the derivatives of y. So $$4y^3 = 12y^2(dy/dx)$$

So the left side of the equation is $$6x^2 + 12y^2(dy/dx)$$

Now we evaluate the right side, We have to use power rule for 12xy because we are multiplying 2 different variables
the formula for power rule is 
$$dy/dx (xy) = x(y') + y(x')$$
so we take x and multiply by the derivative of y, (dy/dx) and add to y times the derivative of x, (dx of dx is 1) so we would have $$12x(dy/dx) + 12y$$
All in all, we have
 $$6x^2 + 12y^2(dy/dx) = 12x(dy/dx) + 12y$$Now we need to get the terms that have dy/dx on one side, so I'll just subtract both sides by  $$12x(dy/dx) $$
$$6x^2 + 12y^2(dy/dx) -12x(dy/dx) = 12y$$
We also subtract both sides by 6x^2.
$$12y^2(dy/dx) -12x(dy/dx) = 12y-6x^2$$
Factor out dy/dx
$$(dy/dx)(12y^2 -12x)= 12y-6x^2$$
Simplifying we have
$$(dy/dx) = ( 12y-6x^2)/(12y^2 -12x)=$$
Which can be further simplified into $$dy/dx = (2y-x^2)/(2(y^2-x))$$

darkknight · Answer

Also btw, you need a background in Calculus. Like you need to know power, constant and quotient rules for derivatives beforehand. I just made this so that some one who needs help can reference this topic