What is the derivative of 4^x ?

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

What is the first derivative of 4^x ?

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

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derivative of

4^{x}

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

Solution steps

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

Apply the derivative exponent rule:

\frac{d}{d x} (a^{x}) = a^{x} ln (a)

= 4^{x} ln (4)

Simplify

4^{x} ln (4) : ln (2) \cdot 2^{2 x + 1}

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

\int_{0}^{4} x e^{- 3 x} d x

\frac{\partial}{\partial x} (v sin (x))

f (x) = 2 + x

\int e^{- 5 x^{2}} d x

\frac{\partial}{\partial x} (1 - \frac{x}{y})

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