What is the derivative of 3^{1-3x} ?

The derivative of 3^{1-3x} is -ln(3)*3^{-3x+2}

What is the first derivative of 3^{1-3x} ?

The first derivative of 3^{1-3x} is -ln(3)*3^{-3x+2}

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

- ln (3) \cdot 3^{- 3 x + 2}

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

3^{1 - 3 x} = e^{(1 - 3 x) l n (3)}

= \frac{d}{d x} (e^{(1 - 3 x) l n (3)})

Apply the chain rule:

e^{(1 - 3 x) l n (3)} \frac{d}{d x} ((1 - 3 x) ln (3))

= e^{(1 - 3 x) l n (3)} \frac{d}{d x} ((1 - 3 x) ln (3))

\frac{d}{d x} ((1 - 3 x) ln (3)) = - 3 ln (3)

= e^{(1 - 3 x) l n (3)} (- 3 ln (3))

Simplify

e^{(1 - 3 x) l n (3)} (- 3 ln (3)) : - ln (3) \cdot 3^{- 3 x + 2}

= - ln (3) \cdot 3^{- 3 x + 2}

\int e^{- s t} (t sin (t)) d t

\int sin^{2} (t) cos^{4} (t) d t

y = 3 x + 4

\frac{d}{d x} (\frac{2 x}{( x + 1 )})

2 x + 3

What is the derivative of 3^{1-3x} ?
The derivative of 3^{1-3x} is -ln(3)*3^{-3x+2}
What is the first derivative of 3^{1-3x} ?
The first derivative of 3^{1-3x} is -ln(3)*3^{-3x+2}