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