Question

What is the difference between implicit and explicit differentiation?

Answer 1

Hint: Here in this question, we have to explain the difference between the explicit and implicit differentiation. For this first we have to know what is implicit differentiation and explicit differentiation and later discuss their difference between one by one.

Complete step by step solution:
Explicit Differentiation is taking the derivative of an explicit function. In an explicit function, one variable is defined completely in terms of the other. This usually means that the independent variable (x) is written explicitly in terms of the dependent variable (y).
The general form is: \[y = f\left( x \right)\] .
Note that “y” is on one side of the equals sign and “x” is on the other side.
Most of the functions you’re probably familiar with are explicit, like \[y = {x^2}\] or \[y = 2x + 3\] . . When you first start in calculus, practically all of the functions you work with are going to be in this explicit form, and you’ll use the usual rules for differentiation.
Implicit differentiation, if you ask me, is slightly confusingly named. The "implicit" does not refer to the act of differentiation, but to the function being differentiated.
Implicit differentiation means "differentiating an implicitly defined function".
The opposite of an explicit function is an implicit function, where the variables become a little more muddled. For example, the following equations are implicit:
 \[{x^2} + {y^2} = 1\] (x and y are on one side of the equation)
 \[y \cdot {e^y} = x\] (two “y” s are on one side of the equation).
Example:
Let’s say you wanted to differentiate the implicit function \[{x^2} + {y^2} = 4\]
Using explicit functions.
Rewrite, using algebra, so that you have one variable on each side of the equals sign:
 \[y = f\left( x \right) = \pm \sqrt {4 - {x^2}} \]
This would give you two derivatives, one for positive values of y, and one for negative values of y.
 Using implicit differentiation:
Instead of rewriting, you can just go ahead and plug the function in:
 \[2x + 2y\dfrac{{dy}}{{dx}} = 0\]
 \[ \Rightarrow \,\,\dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{{2y}}\]
 \[ \Rightarrow \,\,\dfrac{{dy}}{{dx}} = - \dfrac{x}{y}\]

Note: When "regular" differentiation, we can write one variable as a function of the other and then just go to town taking the derivatives. But what happens when you cannot express one as a function of the other? That is where the technique of implicit differentiation comes in. It does just get you the derivative, but for expressions that are difficult to solve "regularly".