Implicit and Logarithmic Differentiation

When we're presented with a function such as y = 5x 2 + 3x + 1, we say that the function is explicitly defined. We mean that y is a function of x. To find , we apply the power rule and we're done. Even if we're given x 2 + y 2 = 25, we can find the explicit function y by solving the equation like this: now we can find by applying the chain rule.

But when we have to find or y ' for a mess like x 2 y 7 –5x 3 y 2 –17y 5 = 29, it could take us weeks to find the explicit function y = f (x). This is when we use implicit differentiation.

Here's the theory. We assume that there is a function y = f (x) and we want its derivative. So, we apply the proper rules of differentiation to the expression -- and as we do --

when we differentiate the y-terms in the expression

we write y ' -- the thing we're looking for.

To prove that the results are the same explicitly and implicitly, let's use which represents a circle, center (0, 0) with radius = 5.

First, we'll differentiate it explicitly, then implicitly.

 explicitly:   implicitly x 2 + y 2 = 25 2x + 2y y' = 0 (divide by 2, transpose) , the derivatives are the same.

Note: when we differentiate y²-- we get 2 y y ' -- since we're differentiating a y-term.

Rule: when differentiating implicitly --

we put in y ' any time we differentiate a y-term.

Now an example with products -- like x 2 . y 7 and 5x 3 . y 2

mark them (.) to remind yourself.

Example: given x 2.y 7 – 5x 3.y 2 – 17y 5 = 29, find y ' .

We differentiate each term, putting in the y ' where needed.

Since the first 2 terms are products, we use product rule.

2x y 7 + 7x 2 y 6 y ' – 15x 2 y 2 – 10x 3 y y ' – 85y 4 y ' = 0

factor out y ' on the left -- transpose everybody else --

y ' ( 7x 2 y 6 – 10x 3 y – 85y 4 ) = 15x 2 y 2 – 2x y 7

then divide. With implicit differentiation, we need both coordinates for the point of contact of the tangent to evaluate the derivative.

Example: Find the equation of the tangent to x 2.y 7 – 5x 3.y 2 – 17y 5 = – 21 at P (1, 1)

from , setting x = 1 and y = 1

we get y ' or the slope of the tangent is So, the equation of the tangent at (1, 1) is , or One more example with a quotient -- yuck! -- I hate these -- especially if we need to find the 2nd derivative -- so we'll keep it simple. A good way around the quotient rule is to multiply through by the common denominator(s) and turn the expression into a product if it is not too complicated.

Example: Differentiate  we break up the fraction like this factor out y ' on the left -- transpose everybody else -- Now we divide by the bracket on the left to get a royal mess of a fraction.

Better, if possible, to multiply through by y before differentiating or y 2 after.

Let's do it again sanely -- we'll multiply through by y to eliminate the fraction before differentiating. differentiating -- pay attention to the product --

2xy 4 + 4x 2 y 3 y ' 5 + 9y 2 y' 9y' = 0

y' (4x 2 y 3 + 9y 2 9 ) = 5 – 2xy 4 .

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(otherwise known as log diff)

There are 2 situations in which log diff is used. In the first, it is merely useful for simplifying an ugly mess of function before differentiating it. In the second -- log diff is the only way to handle the question.

Case 1: when dealing with a mess of a function. Since the rules of logs will change a power into a product, and a quotient into a difference, taking natural log of this mess will simplify the differentiating.

First, we must name the thing y, then take natural log of both sides, like this:

Differentiate Name it y -- take ln of both sides -- apply log rules.  Now we multiply through by y in its original form, so .

Case 2: differentiating a variable to a variable power.

So far, all our differentiation rules apply to either a variable to a constant power such as x 5 or an exponential function which is a constant to a variable power -- like (5) 2 x – 3 -- but what to do with a function like f (x) = x sin x ?? Both the base and the exponent are variables.

Again, we log both sides of the equation, apply the rules of logs and differentiate. At the end, we mustn't forget to multiply through by y in its original form.

Let's do y = x sin x as our example.

ln y = (sin x)(ln x)  One more example for those who don't do trig functions (commerce program)

Find y ' if y = (2x + 3) 7 – 5 x

ln y = (7 – 5x) ln(2x + 3)  In both cases of log diff -- we mustn't forget that we want y ' and not .

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1) Write the equation of the tangent to the curve x 2 y + 3xy2 – 2 y 3 + 2 = 0 at P(1, 2)

2) Find y ' if :

a) x sin y = y cos x
b) y = ( x + 7 ) 5 x – 3
c) y 2 e 5 x + xy 3 = 1
d) 3) Use logarithmic differentiation to find y ' if  1) Write the equation of the tangent to the curve x 2 y + 3xy2 – 2 y 3 + 2 = 0 at P(1, 2)

2xy + x 2 y' + 3y 2 + 6xyy' – 6y 2 y' = 0
y ' (x 2 + 6xy – 6y 2 ) = – 2xy – 3y 2  for the equation of the tangent.

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2) Find y ' if :

 a) x sin y = y cos x sin y + xy ' cos y = y ' cos x – y sin x y ' ( x cos y –cos x ) = – sin y – y sin x b) y = ( x + 7 ) 5 x – 3 ln y = (5x – 3) ln (x + 7)  c) y 2 e 5 x + xy 3 = 1 2e 5 x yy ' + 5e 5 x y 2 + y 3 + 3xy 2 y ' = 0 y ' (2e 5 x y + 3xy 2 ) = – 5e 5 x y 2 – y 3 d) ln y = (x 2 – 1) ln (x + 4)  3) Use logarithmic differentiation to find y ' if    . (all content © MathRoom Learning Service; 2004 - ).