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

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

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

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

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derivative of

2^{x - 1}

ln (2) \cdot 2^{x - 1}

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

x \frac{d y}{d x} + 5 y = x^{3} - x

7 e^{x}, 7 x e^{x^{2}}, 0, 1

\frac{d}{d x} (2 x^{4} - 5 x^{3} + 4)

h \to 0 lim (- 2 x - h + 3)

\frac{d}{d x} (3 cos (4 x) + sin (4 x))

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