*Notes to a video lecture on http://www.unizor.com*

__Implicit Differentiation__

The method of

*implicit differentiation*can be used to find a derivative of a function, implicitly defined by an equation that contains both argument and function value, for example

*x² + y² = R²*where

**is a function of**

*y***.**

*x*In some cases (like the one above) we can start from resolving the given equation for

**as an**

*y**explicit*formula (like

**)**

*y = ±√R²−x²*In other cases it might be impossible or impractical to resolve it for

**, and these are the cases where**

*y**implicit differentiation*would be useful.

Consider the following problem, from which the method will be clear.

*Problem 1*

The function is given by the following implicit equation

*x² + y² = sin(y)*that cannot be explicitly resolved for

**.**

*y*Our purpose is to express the derivative

*dy/dx*in terms of

**and**

*x***.**

*y**Solution*

Assuming that

**is some unknown function of**

*y***, we consider the defining equation as the equality between two functions:**

*x*

*x² + y²***.**

*sin(y)*Since these two functions are equal, their derivatives must be equal as well:

**[**

*x² + y²*]*[*^{I}=*sin(y)*]^{I}Using the property of derivative of a sum and the chain rule for compound functions, this produces:

**[**

*x²*]*[*^{I}+*y²*]*[*^{I}=*sin(y)*]^{I}

*2·x + 2·y·y*^{I}= cos(y)·y^{I}This can be resolved for

**to get its expression in terms of**

*y*^{I}**and**

*x***:**

*y*

*y*^{I}= 2x/(cos(y)−2y)Let's exemplify the method of

*implicit differentiation*further.

Assume that we don't know how to differentiate a logarithmic function. Consider then the following problem.

*Problem 2*

Find the derivative of a function

**y = ln(x)***Solution*

We know the definition of logarithmic function:

*means that*

**y = ln(x)**

*x = e*^{ y}On the left of the last equality is a function

**and on the right - a compound function**

*f(x)=x***, where**

*g(x)=e*^{ y}**, derivative of which we want to calculate.**

*y=ln(x)*Let's differentiate both sides of the above expression using the chain rule for compound function

**:**

*e*^{y}

*D*_{x}(x) = D_{x}(e^{ y})

*1 = e*^{ y}·D_{x}(y)Since

**by definition of logarithmic function**

*e*^{ y}= x*, this results in*

**y = ln(x)**

*1 = x·D*_{x}(y)from which follows

*D*_{x}(y) = 1/xHence,

*D*_{x}(ln(x)) = 1/xwhich is the same as we derived when calculated this derivative directly using the limits.

*Problem 3*

Find the derivative of a function

**y = x**^{sin(x)}*Solution*

*ln(y) = sin(x)·ln(x)*

*D*_{x}(ln(y)) = D_{x}(sin(x)·ln(x))

*(1/y)·D*_{x}(y) = cos(x)·ln(x) + sin(x)·(1/x)

*D*_{x}(y) = y·[cos(x)·ln(x)+sin(x)·(1/x)]Hence, derivative of

**equals to**

*x*^{sin(x)}

*x*^{sin(x)}·[cos(x)·ln(x)+sin(x)·(1/x)]
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