![{x}^{2} + \sqrt{x} + \sqrt[5]{x}](/tpl/images/2409/1215/c356f.png)
Answers: 3
![3 + \frac{1}{2\sqrt{3} } + \frac{1}{5\sqrt[5]{81} }](/tpl/images/1379/7469/6d15c.png)
and ![\sqrt[5]{x} = x^{\frac{1}{5} }](/tpl/images/1379/7469/179af.png)
This way we could more easily use the power rule of derivatives.
then f'(x) will be as follows.![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} }](/tpl/images/1379/7469/1efe5.png)
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what is f'(3) of this equation?...
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