What is (df)/(dt)(2te^{-t^2}) ?

The solution to (df)/(dt)(2te^{-t^2}) is (2t\frac{df)/(dt)}{e^{t^2}}

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simplify (df)/(dt)(2te^{-t^2})

simplify

\frac{df}{d t} (2 t e^{- t^{2}})

\frac{2 t \frac{df}{d t}}{e ^{t^{2}}}

Solution steps

\frac{df}{d t} (2 t e^{- t^{2}})

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= 2 \cdot \frac{1}{e ^{t^{2}}} t \frac{df}{d t}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \frac{df}{d t} 2 t}{e ^{t^{2}}}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2 t \frac{df}{d t}}{e ^{t^{2}}}

34.416 e^{- 0.18 \cdot 6}

x^{8} e^{- x}

e^{- π t^{2}}

e^{- x} \times x

What is (df)/(dt)(2te^{-t^2}) ?
The solution to (df)/(dt)(2te^{-t^2}) is (2t\frac{df)/(dt)}{e^{t^2}}