Derivative

Question

Solve for derivative and find inflection points of f(x)= x (x+3) ^(1/2)....Please go step by step of simplifying derivative because I struggle on that involving square roots.

Answer 1

Hi Kayla.

 

You have here a derivative involving the product rule. The product rule states:

 

If

 

f(x) = g(x)*h(x)

 

then

 

f'(x) = g'(x)*h(x) + g(x)*h'(x)

 

In this scenario, you have

 

g(x) = x

h(x) = (x+3)^1/2

 

Find each derivative:

 

g'(x) = 1

h'(x) = (1/2)*(x+3)^-1/2

 

NOTE: for a function involving square roots (or any root for that matter) all you do is apply the power rule like you do with any other power function (like x² or x³ or x^1/2)

 

As you know, if

 

f(x) = xⁿ

 

then

 

f'(x) = n*x^n-1

 

In the case of the square root, n = 1/2, so

 

f'(x) = 1/2*x^{1/2 - 1}

 

= 1/2*x^-1/2

 

You don't have x, though, you have (x+3). But it's the same derivative since the chain rule says to multiply it by 1.

 

Anyways.

 

f'(x) = g'(x)*h(x) + g(x)*h'(x)

 

and

 

g(x) = x
h(x) = (x+3)^1/2

g'(x) = 1
h'(x) = (1/2)*(x+3)^-1/2

 

So just substitute everything.

 

f'(x) = (1)*(x+3)^1/2 + (x)*((1/2)*(x+3)^-1/2)

 

= (x+3)^1/2 + (x/2)*(x+3)^-1/2

 

Hope this helps.