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