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?
That means take d/dx of everything
x^4 + y^4 = 16
d/dx x^4 = 4 x^3 dx/dx = 4 x^3
d/dx y^4 = 4 y^3 dy/dx
d/dx 16 = 0
so
4 x^3 + 4 y^3 dy/dx = 0
dy/dx = -x^3/y^3
basic thing is
d/dx u^n = n u^(n-1) du/dx
when u happens to be x, you get dx/dx at the end which is 1
but if u happens to be y, then you have that dy/dx at the end
Implicit differentiation is a method used to differentiate both sides of an equation with respect to a common variable, usually "x". It is often used when it is not possible or convenient to express "y" as an explicit function of "x".
When we say "we differentiate both sides of the equation with respect to x," it means that we apply the derivative operator to each term of the equation, treating "x" as the independent variable and "y" as a function dependent on "x" (meaning that "y" is a function of "x").
To understand why we introduce "y'" in the derivative of y, let's recall the chain rule of differentiation. The chain rule states that if "y" depends on "x", then the derivative of "y" with respect to "x" (denoted as "dy/dx" or "y'") is given by the product of the derivative of "y" with respect to "x" and the derivative of "x" with respect to "x" (which is 1). In this context, "y'" represents the derivative of "y" with respect to "x".
In your example, the equation is "x^4 + y^4 = 16". To find "y'", we differentiate both sides of the equation with respect to "x".
Differentiating the left side with respect to "x" is straightforward. The derivative of "x^4" with respect to "x" is "4x^3", and the derivative of a constant (in this case, 16) is zero.
When differentiating the right side of the equation, we have to consider that "y" is a function of "x". The derivative of "y^4" with respect to "x" using the chain rule is 4y^3 * y'. Here, "4y^3" is the derivative of "y^4" with respect to "y" (using the power rule), and "y'" indicates that we are taking the derivative of "y" with respect to "x".
Hence, differentiating both sides of the equation yields:
4x^3 + 4y^3y' = 0.
This equation allows us to find the derivative of "y" (y') in terms of "x" and "y".