What is (\partial)/(\partial t)(te^{-2x}+cos(x)) ?

The answer to (\partial)/(\partial t)(te^{-2x}+cos(x)) is e^{-2x}

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(\partial)/(\partial t)(te^{-2x}+cos(x))

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

e^{- 2 x}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial t} (t e^{- 2 x}) = e^{- 2 x}

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

= e^{- 2 x} + 0

Simplify

= e^{- 2 x}

f (x) = x^{2}, at x = 1.9

\int \frac{1 - 0.4 x}{( 1 - x ) ( 5 - x )} d x

\int_{- \infty}^{\infty} t d t

\frac{d}{d x} ((5 - (4 - 3 x)^{2})^{3})

\frac{d y}{d x} = a^{x + y}

What is (\partial)/(\partial t)(te^{-2x}+cos(x)) ?
The answer to (\partial)/(\partial t)(te^{-2x}+cos(x)) is e^{-2x}