How do you find the derivative of $f(x)=(5x^6sqrt x) + (3/(x^3 sqrt x))$ ?

Question

2228 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-22T01:33:50

$f'(x) = 1/2(65x^5sqrt(x) -21/(x^4sqrt(x)))$

$f(x) = (5x^6sqrt(x)) + (3/(x^3sqrt(x)))$

Using the rules of indicies $f(x)$ can be written:

$f(x) =5x^6x^(1/2) + 3x^-3x^(-1/2)$ $= 5x^(13/2) + 3x^(-7/2)$

Aplying the Power Rule to both terms:

$f'(x) = 5* 13/2 x^(13/2-1) + 3* (-7/2) x^(-7/2-1)$
$= 1/2(65x^(11/2) -21x^(-9/2))$

To express $f'(x)$ in the form of $f(x)$ in the original question, we can rewrite $f'(x)$ as:
$f'(x) = 1/2(65x^5 * x^(1/2) - 21x^(-4) * x^(-1/2))$

$=1/2(65x^5 sqrt(x) - 21/(x^4 sqrt(x)))$

How do you find the derivative of f(x)=(5x^6sqrt x) + (3/(x^3 sqrt x))f(x)=(5x^6sqrt x) + (3/(x^3 sqrt x))?