What is the derivative of 3^{1-2n} ?

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

What is the first derivative of 3^{1-2n} ?

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

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derivative of

3^{1 - 2 n}

- 2 ln (3) \cdot 3^{1 - 2 n}

Solution steps

\frac{d}{d n} (3^{1 - 2 n})

Apply exponent rule:

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

3^{1 - 2 n} = e^{(1 - 2 n) l n (3)}

= \frac{d}{d n} (e^{(1 - 2 n) l n (3)})

Apply the chain rule:

e^{(1 - 2 n) l n (3)} \frac{d}{d n} ((1 - 2 n) ln (3))

= e^{(1 - 2 n) l n (3)} \frac{d}{d n} ((1 - 2 n) ln (3))

\frac{d}{d n} ((1 - 2 n) ln (3)) = - 2 ln (3)

= e^{(1 - 2 n) l n (3)} (- 2 ln (3))

Simplify

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

= - 2 ln (3) \cdot 3^{1 - 2 n}

\frac{\partial}{\partial x} (3 x y - x^{2} y^{2} + 23)

\frac{d y}{d x} = \frac{x}{e ^{y}}

3 x^{2} - 2

f (x) = - 9.8 x^{2} + 2 x + 1

\int \frac{1}{arcsin ( x ) 1 - x ^{2}} d x

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