What is the Explanation for Implicit Differentiation Equation?

In summary, the conversation discusses the equation (d/dx)f=(∂f/∂x)+(∂f/∂y)*(dy/dx) and provides a detailed explanation for it. The equation is only true if f is a function of two variables and there is a function g of one variable such that y=g(x). The conversation then goes on to explain the various terms and limits involved in the equation, ultimately arriving at the final result of (d/dx)f=(∂f/∂x)+(∂f/∂y)*(dy/dx).
  • #1
miguel hernandez
2
0
First of all thanks for the help, i have a problem finding a good explanation of de ecuation (d/dx)f=(∂f/∂x)+(∂f/∂y)*(dy/dx) could anyone write me a good explanation of this ecuation? thanks for the help
 
Physics news on Phys.org
  • #2
This is only true if ##f## is a function of two variables and there is a function ##g## of one variable such that ##y=g(x)##.

Then we have ##f(x,y)=f(x,g(x))## so then, using the definitions of derivatives as limits, we can write:
\begin{align*}
\frac d{dx}f(x,y)&=\frac d{dx}f(x,g(x))\\
&=\lim_{h\to 0}\left[\frac{f(x+h,g(x+h))-f(x,g(x))}{h}\right]\\
&=\lim_{h\to 0}\left[
\frac{f(x+h,g(x+h))-f(x,g(x+h))}{h}
+\frac{f(x,g(x+h))-f(x,g(x))}{h}
\right]\\
&=\lim_{h\to 0}\left[
\frac{f(x+h,g(x+h))-f(x,g(x+h))}{h}
\right]+
\lim_{h\to 0}\left[
\frac{f(x,g(x+h))-f(x,g(x))}{h}
\right]
\end{align*}
provided both limits exist.

The first term is a bit tricky. Provided certain assumptions hold, the details of which we won't go into here, it is equal to the double limit:
\begin{align*}
\lim_{h'\to 0}\bigg[
\lim_{h\to 0}\bigg(
&\frac{f(x+h,g(x+h'))-f(x,g(x+h'))}{h}
\bigg)
\bigg]\\
&=\lim_{h'\to 0}\bigg[
\frac{\partial}{\partial x}f(x,g(x+h'))
\bigg]\\
&=
\frac{\partial}{\partial x}f(x,g(x))\\
\end{align*}

The second term is equal to:
\begin{align*}
\lim_{h\to 0}\bigg[
\frac{f(x,g(x+h))-f(x,g(x))}{g(x+h)-g(x)}
&\times
\frac{g(x+h)g(x)}{h}
\bigg]\\
&=\lim_{h\to 0}\left[
\frac{f(x,g(x+h))-f(x,g(x))}{g(x+h)-g(x)}\right]
\times
\lim_{h\to 0}\left[\frac{g(x+h)-g(x)}{h}
\right]\\
&=\lim_{\delta y\to 0}\left[
\frac{f(x,y+\delta y)-f(x,y)}{\delta y}\right]
\times
\lim_{h\to 0}\left[\frac{g(x+h)-g(x)}{h}
\right]\\
&=
\frac{\partial }{\partial y}f(x,y)
\times
\frac{dg}{dx}(x)\\
&=
\frac{\partial }{\partial y}f(x,y)
\times
\frac{dy}{dx}\\
\end{align*}
Summing the terms, we get
$$\frac{d}{dx}f(x,y)=\frac{\partial}{\partial x}f(x,y)+\frac{\partial }{\partial y}f(x,y)
\times
\frac{dy}{dx}$$
as required.
 
  • Like
Likes miguel hernandez and vanhees71
  • #3
thanks a lot for the help
 

1. What is implicit differentiation?

Implicit differentiation is a mathematical concept used to find the derivative of an equation that is not explicitly expressed in terms of a dependent variable. It is commonly used when an equation cannot be easily solved for the dependent variable, and involves differentiating both sides of the equation with respect to the independent variable.

2. When is implicit differentiation used?

Implicit differentiation is used when an equation is not in the form of y = f(x), meaning the dependent variable is not explicitly expressed in terms of the independent variable. It is commonly used in multivariable calculus, differential equations, and physics problems.

3. What is the process of implicit differentiation?

The process of implicit differentiation involves differentiating both sides of an equation with respect to the independent variable, treating the dependent variable as a function of the independent variable. This will result in an equation with the derivative of the dependent variable in terms of the derivative of the independent variable.

4. What is the difference between implicit and explicit differentiation?

Explicit differentiation involves finding the derivative of an equation explicitly expressed in terms of the independent variable, while implicit differentiation involves finding the derivative of an equation that is not explicitly expressed in terms of the dependent variable. In other words, implicit differentiation is used for equations that cannot be easily solved for the dependent variable.

5. What are some real-world applications of implicit differentiation?

Some real-world applications of implicit differentiation include optimization problems in economics and physics, determining the rate of change in a chemical reaction, and finding the velocity and acceleration of a moving object in physics.

Similar threads

Replies
3
Views
1K
  • Calculus
Replies
16
Views
2K
Replies
9
Views
1K
Replies
2
Views
2K
Replies
14
Views
1K
Replies
1
Views
954
  • Calculus
Replies
8
Views
1K
Replies
9
Views
2K
  • Calculus
Replies
25
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
728
Back
Top