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

\int (x - \frac{1}{2} x + \frac{2}{x}) d x

x^{- \frac{3}{5}} + x^{\frac{3}{5}}

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

\int_{\frac{1}{2}}^{7} 16 x ln (2 x) d x

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}