Quotient Rule Example

Q:  Find dy/dx of

y = x / sqrt(x2 + 1)

A:  To find dy/dx (the derivative), we will need to use the quotient rule since we have a function over a function.  See? We are in the form:

y = f / g where f and g are two different functions of x.

In this form, the quotient rule tells us if:

y = f / g
then

y ‘ = (g * f ‘ – f * g ‘ ) / g2

So, we know:

f = x

g = sqrt(x2 + 1)

f ‘ = 1   <— basic derivative

g ‘ = (this takes more work…. First, rewrite “g”):

g = sqrt(x2 + 1) = (x2 + 1)1/2

Now, use a power rule and a chain rule to find g ‘ like so:

g ‘ = 1/2 (x2 + 1) – 1/2 (2x)

[that’s the derivative of the outside * the derivative of the inside]

Clean up g ‘:

g ‘ = x (x2 + 1) – 1/2

So now we have all of the players:  f, f ‘ , g, g ‘:

f = x

g = (x2 + 1)1/2

f ‘ = 1

g ‘ = x (x2 + 1) – 1/2

Now, we said:  y ‘ = (g * f ‘ – f * g ‘ ) / g2, so plug in the pieces then simplify like so:

y ‘ = [ (x2 + 1)1/2(1) – (x)(x (x2 + 1) – 1/2 ) ] / [(x2 + 1)1/2]2

CLEAN UP

y ‘ = [ (x2 + 1)1/2 – x2(x2 + 1) – 1/2  ] / (x2 + 1)

Multiply top and bottom by (x2 + 1)1/2

[ (x2 + 1)1/2(x2 + 1)1/2 (x2 + 1)1/2x2(x2 + 1) – 1/2  ] / (x2 + 1)1/2(x2 + 1) =

[ (x2 + 1)- x2 ] / (x2 + 1)3/2

1 / (x2 + 1)3/2

TADA!

So, if:

y = x / sqrt(x2 + 1)

dy/dx = 1 / (x2 + 1)3/2

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