What is (\partial)/(\partial x)(2cos(t)+8x^2) ?

The answer to (\partial)/(\partial x)(2cos(t)+8x^2) is 16x

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(\partial)/(\partial x)(2cos(t)+8x^2)

\frac{\partial}{\partial x} (2 cos (t) + 8 x^{2})

16 x

Solution steps

\frac{\partial}{\partial x} (2 cos (t) + 8 x^{2})

Treat

t

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (2 cos (t)) + \frac{\partial}{\partial x} (8 x^{2})

\frac{\partial}{\partial x} (2 cos (t)) = 0

\frac{\partial}{\partial x} (8 x^{2}) = 16 x

= 0 + 16 x

Simplify

= 16 x

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What is (\partial)/(\partial x)(2cos(t)+8x^2) ?
The answer to (\partial)/(\partial x)(2cos(t)+8x^2) is 16x