Can someone explain the method of implicit differentiation to me.

I know that we differentiation both sides of the equation with respect to x (I don't understand what they mean by with respect to x). And then y is a function of x...I don't understand what that means.

For example,
Find y'' if x^4+y^4 = 16

Basically we need to differentiate both sides..as far as I know..

so ..(for y')
4x^3+4y^3 = 0

BUT..
in the textbook it's

4x^3+4y^3y' = 0

Where does the y' come from?

Implicit differentiation is a technique used to differentiate equations that are not explicitly written in terms of one variable, usually x. It allows us to find the derivative of the dependent variable, y, with respect to the independent variable, x.

To use implicit differentiation, you differentiate both sides of the equation with respect to x. When we say "with respect to x," it means that we are treating x as the variable we are differentiating with respect to.

In your example, the equation is x^4 + y^4 = 16. By differentiating both sides of the equation with respect to x, we treat x as the variable and differentiate each term individually.

For the left side, the derivative of x^4 with respect to x is 4x^3, which is straightforward because x^4 is written explicitly in terms of x.

However, for the right side, y^4 is written in terms of y, not x. This is where implicit differentiation comes in. Since y is a function of x, we treat it as an implicit function and apply the chain rule.

When we differentiate y^4 with respect to x, we need to consider the derivative of an implicit function, which involves taking the derivative of y with respect to x (y') and then multiplying it by the derivative of y^4 with respect to y. The derivative of y^4 with respect to y is 4y^3.

So, differentiating the equation x^4 + y^4 = 16 with respect to x results in:

4x^3 + 4y^3 * y' = 0

This is why the derivative of y^4 is 4y^3y'.

Now, to find y'' (the second derivative of y with respect to x), you need to differentiate the equation 4x^3 + 4y^3 * y' = 0 with respect to x again. This time, you will be differentiating y' (the derivative of y with respect to x) as well.

I hope this explanation helps clarify the steps involved in using implicit differentiation!