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

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

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

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

- 2 sin (t)

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial t} (2 cos (t)) = - 2 sin (t)

\frac{\partial}{\partial t} (8 x^{2}) = 0

= - 2 sin (t) + 0

Simplify

= - 2 sin (t)

\frac{d}{d x} (3 cos (x) - 2 e^{2 x})

f (x) = x^{3} - 2 x + 1, at x = - 1

(e^{x} - e^{- x}) \frac{d y}{d x} = y^{2}

\frac{\partial}{\partial x} (- 10 x + 7 y^{5})

\frac{d}{d x} (\frac{1}{4 x ^{3}} - \frac{6}{5 x ^{4}})

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