Friday, April 27, 2012

Calculus Revisited #6: The Chain Rule

Today we cover a very useful rule in Calculus: The Chain Rule. The rule deals with so many applications in higher mathematics. This is entry #6 in a series of 21.

The Chain Rule
d/dx f(g(x)) = f'(g(x)) * g'(x) = df/dg * dg/dx

If you learn nothing else in calculus - learn this rule! We will do some practice problems with chain rule today.

Problems
1. d/dx (3x^2 + 1)^2

Using the rule above, let f = g^2 and g = 3x^2 + 1.

Then d/dx f(g) = df/dg * dg/dx,
f = g^2
df/dg = 2g
g = 3x^2 + 1
dg/dx = 6x

Putting it all together:
d/dx (3x^2 + 1)^2 = 2g * 6x = 2(3x^2 + 1) * 6x = 36x^3 + 12x

2. d/dx cos(2x^3 - 1)

Let f = cos g and g = 2x^3 -1
Then:
df/dg = -sin g
dg/dx = 6x^2

Put it together:
d/dx cos(2x^3 - 1) = -sin g * 6x^2 = -sin(2x^3 -1) * 6x^2 = -6x^2 * sin(2x^3 - 1)

3. d/dx √(e^x + 1)

Note √x = x^(1/2)

Let f = √g and g = e^x + 1
Then:
df/dg = 1/2 * 1/√g
dg/dx = e^x

So:
d/dx √(e^x + 1) = 1/2 * 1/√g * e^x = e^x / (2 * √(e^x + 1))

Now that you get the though process, let's try another one.

4. d/dx cos^3 x = d/dx (cos x)^3

d/dx (cos x)^3 = 3 * (cos x)^2 * d/dx cos x = 3 * (cos x)^2 * -sin x = - 3 cos^2 x sin x

5. Evaluate the slope of the function y(x) = ln sin x at x = π/4

d/dx ln sin x
= 1 / sin x * d/dx (sin x)
= 1 / sin x * cos x
= cot x

To find the slope at x = π/4, find cot π/4 = 1.

6. The Chain Rule can be applied more than once. This problem illustrates it. Find d/dx ln (cos √x)

d/dx ln (cos √x)
= 1/(cos √x) * d/dx (cos √x)
= 1/(cos √x) * (-sin √x) * d/dx (√x)
= (-sin √x)/(cos √x) * 1/(2√x)
= (-tan √x)/(2√x)

Next time, we will find the extremes of functions.

This blog is property of Edward Shore. © 2012

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