What is the derivative of 4^{x+1} ?

The derivative of 4^{x+1} is ln(2)*2^{2x+3}

What is the first derivative of 4^{x+1} ?

The first derivative of 4^{x+1} is ln(2)*2^{2x+3}

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derivative of

4^{x + 1}

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

Solution steps

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

Apply exponent rule:

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

4^{x + 1} = e^{(x + 1) l n (4)}

= \frac{d}{d x} (e^{(x + 1) l n (4)})

Apply the chain rule:

e^{(x + 1) l n (4)} \frac{d}{d x} ((x + 1) ln (4))

= e^{(x + 1) l n (4)} \frac{d}{d x} ((x + 1) ln (4))

\frac{d}{d x} ((x + 1) ln (4)) = 2 ln (2)

= e^{(x + 1) l n (4)} \cdot 2 ln (2)

Simplify

e^{(x + 1) l n (4)} \cdot 2 ln (2) : ln (2) \cdot 2^{2 x + 3}

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

x \to 1 lim (2 x^{3} - 4 x^{2} + x + 7)

\frac{\partial}{\partial x} (cos (2 y))

\int 39 - 10 x - x^{2} d x

\frac{d}{d x} (18 x - x^{2})

\frac{d}{d x} (4 x + 8)

What is the derivative of 4^{x+1} ?
The derivative of 4^{x+1} is ln(2)*2^{2x+3}
What is the first derivative of 4^{x+1} ?
The first derivative of 4^{x+1} is ln(2)*2^{2x+3}