Question

${x}^{2} + \sqrt{x} + \sqrt[5]{x}$
what is f'(3) of this equation?​

Answer 1

Answer:

$3 + \frac{1}{2\sqrt{3} } + \frac{1}{5\sqrt[5]{81} }$

Step-by-step explanation:

Just to make it easier to see, $\sqrt{x} = x^{\frac{1}{2} }$ and $\sqrt[5]{x} = x^{\frac{1}{5} }$  This way we could more easily use the power rule of derivatives.

So if f(x) = $x^{2} +x^{\frac{1}{2} } +x^{\frac{1}{5} }$ then f'(x) will be as follows.

f'(x) = $x^{1} +\frac{1}{2} x^{-\frac{1}{2} } +\frac{1}{5} x^{-\frac{4}{5} } = x +\frac{1}{2x^{\frac{1}{2} }} +\frac{1}{ 5x^{\frac{4}{5} }} = x +\frac{1}{2\sqrt{x}} +\frac{1}{ 5\sqrt[5]{x^4} }$

to find f'(3) just plug 3 into f'(x) so $3 + \frac{1}{2\sqrt{3} } + \frac{1}{5\sqrt[5]{81} }$