What is the derivative of #y = (x^2+4x+3)/sqrtx#?

What is the derivative of #y = (x^2+4x+3) / sqrtx#

1 Answer
Oct 10, 2017

#y' =3/2 x^(1/2) + 2x^(-1/2) - 3/2x^(-3/2)#

Explanation:

Step 1: Simply the function
Since #sqrtx# is on the denominator, we can think of it as #x^(-1/2)# and multiply through to the numerator:
#y = (x^2+4x+3)x^(-1/2)#

Step 2: Distribution
After simplifying the original equation, we can now distribute #x^(-1/2)# to the numerator #(x^2 + 4x +3)#. We get #y = x^(3/2)+4x^(1/2)+3x^(-1/2)#

Step 3: Take the derivative
We can see now the function is in its simplest form as #y = x^(3/2)+4x^(1/2)+3x^(-1/2)#. We can take the derivative of each term in the function by applying the power rule. #y' = (3/2)x^(1/2)+(1/2)4x^(-1/2)+(-1/2)3x^(-3/2)# And finally, the answer will be #y' =3/2 x^(1/2) + 2x^(-1/2) - 3/2x^(-3/2)#