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

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

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

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

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derivative of

2^{2 - 3 x}

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{( x ^{2} + x - 17 )}{( x - 4 )}

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\frac{d}{d θ} (cos (θ) + i sin (θ))

x \to 0 lim (7)

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