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

The derivative of y=3^{1-x} is -ln(3)*3^{1-x}

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

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

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derivative of

y = 3^{1 - x}

- ln (3) \cdot 3^{1 - x}

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

= - ln (3) \cdot 3^{1 - x}

x, \frac{1}{x}, x = 3

\int sin (2 x + \frac{π}{2}) d x

f (x) = 3 t - t^{2}

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

f (x) = (x^{2} + 1)^{x + 1}, at x = 0

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