How do you find the derivative of # (3x^2 + 3x + 4) / sqrt(x)#?

1 Answer
Mar 22, 2016

#f'(x)=(9x^2+3x-4)/(2x^(3/2))#

Explanation:

The first to step is to rewrite this so that it looks more like a polynomial. Use numerical exponents:

#f(x)=(3x^2+3x^1+4)/x^(1/2)#

Split up the numerator:

#f(x)=(3x^2)/x^(1/2)+(3x^1)/x^(1/2)+4/x^(1/2)#

Divide the terms:

#f(x)=3x^(2-1/2)+3x^(1-1/2)+4x^(-1/2)#

#f(x)=3x^(3/2)+3x^(1/2)+4x^(-1/2)#

Now, differentiate each term individually:

#d/dx(x^n)=nx^(n-1)#

This gives us a derivative of:

#f'(x)=3(3/2)x^(3/2-1)+3(1/2)x^(1/2-1)+4(-1/2)x^(-1/2-1)#

#f'(x)=9/2x^(1/2)+3/2x^(-1/2)-2x^(-3/2)#

This could be your final answer, but it's easier to work with functions like this when they resemble the original function. To do this, multiply #f'(x)# by #(2x^(3/2))/(2x^(3/2))#.

#f'(x)=((9/2x^(1/2)+3/2x^(-1/2)-2x^(-3/2))/1)(2x^(3/2))/(2x^(3/2))#

#f'(x)=(9x^2+3x-4)/(2x^(3/2))#