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

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

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

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

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derivative of

y = 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}

6 \cdot y^{^{''}} + 2 \cdot y^{^{'}} = 0, y^{^{'}} (0) = 4

\int_{0}^{1} \frac{2 ( x + 2 )}{5} d x

tan (2^{x})

\frac{d}{d x} ((7 x^{2} - 11 x) e^{x})

x \to 8 - lim (\frac{2 x}{x + 8})

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