What is the derivative of (t({g}(t,x)e^{-3t})) ?

The derivative of (t({g}(t,x)e^{-3t})) is te^{-3t} derivative of {g}(t,x)

What is the first derivative of (t({g}(t,x)e^{-3t})) ?

The first derivative of (t({g}(t,x)e^{-3t})) is te^{-3t} derivative of {g}(t,x)

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\frac{d}{d x} ((t) (g (t, x) e^{- 3 t}))

t e^{- 3 t} \frac{d}{d x} (g (t, x))

Solution steps

\frac{d}{d x} ((t) (g (t, x) e^{- 3 t}))

Treat

t

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= t e^{- 3 t} \frac{d}{d x} (g (t, x))

Y^{2} X^{2} (ah + 1)

3.58 \cdot 1 0^{- 6}

\frac{- 250}{10}

(2 y^{4}) 52 y^{5}

What is the derivative of (t({g}(t,x)e^{-3t})) ?
The derivative of (t({g}(t,x)e^{-3t})) is te^{-3t} derivative of {g}(t,x)
What is the first derivative of (t({g}(t,x)e^{-3t})) ?
The first derivative of (t({g}(t,x)e^{-3t})) is te^{-3t} derivative of {g}(t,x)