Question

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

Answer 1

`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))`