can someone please help me find the derivative of this:

Expressing this in exponential form, we have :

f (x) = x^1/3 + 2x^-2016 + x^-1/3        use the power rule here

f ' (x)   = (1/3)x^-2/3 - 4032x^-2017 - (1/3)x^-4/3

Find the derivative of the following via implicit differentiation:
d/dx(f(x)) = d/dx(2/x^2016+1/x^(1/3)+x^(1/3))
The derivative of f(x) is f'(x):
f'(x) = d/dx(2/x^2016+1/x^(1/3)+x^(1/3))
Differentiate the sum term by term and factor out constants:
f'(x) = 2 d/dx(1/x^2016)+d/dx(1/x^(1/3))+d/dx(x^(1/3))
Use the power rule, d/dx(x^n) = n x^(n-1), where n = -2016: d/dx(1/x^2016) = d/dx(x^(-2016)) = -2016 x^(-2017):
f'(x) = d/dx(1/x^(1/3))+d/dx(x^(1/3))+2 (-2016)/(x^2017)
Simplify the expression:
f'(x) = -4032/x^2017+d/dx(1/x^(1/3))+d/dx(x^(1/3))
Use the power rule, d/dx(x^n) = n x^(n-1), where n = -1/3: d/dx(1/x^(1/3)) = d/dx(x^(-1/3)) = -1/3 x^(-4/3):
f'(x) = -4032/x^2017+d/dx(x^(1/3))+(-1)/(3 x^(4/3))
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 1/3: d/dx(x^(1/3)) = d/dx(x^(1/3)) = x^(-2/3)/3:
f'(x) = -4032/x^2017-1/(3 x^(4/3))+(1)/(3 x^(2/3))
Expand the left hand side:
Answer: |f'(x) = -4032 / x^2017-1 / (3 x^(4/3)) + 1/(3 x^(2/3))

can someone please help me find the derivative of this:
$f(x)=\sqrt[3]{x}+\frac{2}{{x}^{2016}}+\frac{1}{\sqrt{x^3}}$

$\begin{array}{|rcllll|} \hline f(x) &=& \sqrt[3]{x}+\frac{2}{{x}^{2016}}+\frac{1}{\sqrt{x^3}} \\ && & \sqrt[3]{x} = x^{\frac13} \\ && & \frac{2}{{x}^{2016}} = 2 \cdot x^{-2016} \\ && & \frac{1}{\sqrt{x^3}} = \frac{ 1 } { x^{\frac32} } = x^{-\frac32} \\ f(x) &=& x^{\frac13} + 2 \cdot x^{-2016} + x^{-\frac32} \\ && & \text{Formula: } \\ && & y = x^n \\ && & y' = n\cdot x^{n-1} \\\\ f'(x) &=& \frac13 \cdot x^{\frac13-1} + 2 \cdot (-2016) \cdot x^{-2016-1} + (-\frac32) x^{-\frac32-1} \\ f'(x) &=& \frac13 \cdot x^{-\frac23 } -4032 \cdot x^{-2017} -\frac32\cdot x^{-\frac52} \\ && & x^{-\frac23 } = \frac{ 1 } { x^{\frac23} } = \frac{1}{\sqrt[3]{x^2}} \\ && & x^{-2017} = \frac{1}{{x}^{2017}} \\ && & x^{-\frac52 } = \frac{ 1 } { x^{\frac52} } = \frac{1}{\sqrt{x^5}}\\ f'(x) &=& \frac13 \cdot \frac{1}{\sqrt[3]{x^2}} -4032 \cdot \frac{1}{{x}^{2017}} -\frac32\cdot \frac{1}{\sqrt{x^5}} \\ \mathbf{f'(x)} & \mathbf{=} & \mathbf{\frac{1}{3 \sqrt[3]{x^2}} - \frac{4032}{{x}^{2017}} - \frac{3}{2 \sqrt{x^5}} }\\ \hline \end{array}$

Your answer is the same and also correct.