**Q: Find dy/dx of**

**y = x / sqrt(x ^{2} + 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 ‘ ) / g^{2}

So, we know:

f = x

g = sqrt(x^{2} + 1)

f ‘ = 1 <— basic derivative

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

g = sqrt(x^{2} + 1) = (x^{2} + 1)^{1/2}

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

g ‘ = 1/2 (x^{2} + 1)^{ – 1/2 }(2x)

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

Clean up g ‘:

g ‘ = x (x^{2} + 1)^{ – 1/2 }

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

**f = x**

**g = (x ^{2} + 1)^{1/2}**

**f ‘ = 1**

**g ‘ = x (x ^{2} + 1)^{ – 1/2 }**

Now, we said: y ‘ = (g * f ‘ – f * g ‘ ) / g^{2}, so plug in the pieces then simplify like so:

y ‘ = [ (x^{2} + 1)^{1/2}(1) – (x)(x (x^{2} + 1)^{ – 1/2 }) ] / [(x^{2} + 1)^{1/2}]^{2}

CLEAN UP

y ‘ = [ (x^{2} + 1)^{1/2} – x^{2}(x^{2} + 1)^{ – 1/2 } ] / (x^{2} + 1)

Multiply top and bottom by (x^{2} + 1)^{1/2}

[ **(x ^{2} + 1)^{1/2}**(x

^{2}+ 1)

^{1/2 }–

**(x**x

^{2}+ 1)^{1/2}^{2}(x

^{2}+ 1)

^{ – 1/2 }] /

**(x**(x

^{2}+ 1)^{1/2}^{2}+ 1) =

[ (x^{2} + 1)- x^{2} ] / (x^{2} + 1)^{3/2}

1 / (x^{2} + 1)^{3/2}

TADA!

So, if:

**y = x / sqrt(x ^{2} + 1)**

dy/dx = 1 / (x^{2} + 1)^{3/2}