What is implicit differentiation...as in what would be the steps in solving it. I read my textbook but it seems very confusing.
1) Differentiate both sides with respect to x
2) Realize that y is a function of x
3) Solve for y
Plus the d/dx notation made it even more confusing. Can someone please explain this to me simply? Thanks
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html
That is an excellent discussion.
Here is an example:
x^3 + xy^2 + x^2 y=14 find dy/dx
by implicit diff:
3x^2 dx/dx + y^2 dx/dx + 2yxdy/dx +2xydx/dx+x^2 dy/dx=0
butdx/dx is one, so
3x^2+y^2+2xy + dy/dx (2xy+x^2)=0
so dy/dx=-(3x^2+y^3 +2xy)/(2xy-x^2)
Sure! I'd be happy to explain implicit differentiation in a simplified way.
Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly expressed in terms of x. It is often used when the equation involves both x and y variables and cannot be easily solved for y in terms of x.
Here are the steps to perform implicit differentiation:
Step 1: Differentiate both sides of the equation with respect to x. Treat y as a function of x, even if it is not explicitly written as y = f(x).
Step 2: Apply the chain rule whenever you encounter a term involving y. To do this, multiply the derivative of y with respect to x (dy/dx) by the derivative of the term inside the parentheses.
Step 3: Simplify and collect all terms involving dy/dx on one side of the equation. Then, isolate dy/dx to find its value.
Step 4: If you need to solve for y, you can do so by manipulating the differentiated equation, but keep in mind that finding y might not always be possible or necessary.
To illustrate these steps with an example, let's say we have the equation x^2 + y^2 = 1. We want to find dy/dx.
Step 1: Differentiate both sides of the equation with respect to x:
d/dx (x^2 + y^2) = d/dx(1)
2x + 2yy' = 0, where y' represents dy/dx.
Step 2: Apply the chain rule to the term involving y:
2x + 2y * dy/dx = 0
Step 3: Simplify and isolate dy/dx:
2y * dy/dx = -2x
dy/dx = -2x / 2y
dy/dx = -x/y
So, the derivative of y with respect to x is -x/y.
Remember, implicit differentiation is a useful technique when you have an equation involving both x and y variables and need to find the derivative. It helps us work with functions that are not explicitly given in terms of x.