Monday, December 12, 2016

Unizor - Derivatives - Implicit Differentiation

Notes to a video lecture on

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 y is a function of x.

In some cases (like the one above) we can start from resolving the given equation for y as an explicit formula (like y = ±√R²−x²) and then take a derivative.
In other cases it might be impossible or impractical to resolve it for y, and these are the cases where 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 y.


Assuming that y is some unknown function of x, we consider the defining equation as the equality between two functions: x² + y² and 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:
[]I + []I = [sin(y)]I
2·x + 2·y·yI = cos(y)·yI
This can be resolved for yI to get its expression in terms of and y:
yI = 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)


We know the definition of logarithmic function:
y = ln(x) means that
x = e y

On the left of the last equality is a function f(x)=x and on the right - a compound function g(x)=e y, where y=ln(x), derivative of which we want to calculate.
Let's differentiate both sides of the above expression using the chain rule for compound function ey:
Dx(x) = Dx(e y)
1 = e y·Dx(y)
Since e y = x by definition of logarithmic function y = ln(x), this results in
1 = x·Dx(y)
from which follows
Dx(y) = 1/x

Dx(ln(x)) = 1/x
which is the same as we derived when calculated this derivative directly using the limits.

Problem 3

Find the derivative of a function y = xsin(x)


ln(y) = sin(x)·ln(x)
Dx(ln(y)) = Dx(sin(x)·ln(x))
(1/y)·Dx(y) = cos(x)·ln(x) + sin(x)·(1/x)
Dx(y) = y·[cos(x)·ln(x)+sin(x)·(1/x)]
Hence, derivative of xsin(x)equals to

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