What is the derivative of 4^x-7 ?

The derivative of 4^x-7 is ln(2)*2^{2x+1}

What is the first derivative of 4^x-7 ?

The first derivative of 4^x-7 is ln(2)*2^{2x+1}

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\frac{d}{d x} (4^{x} - 7)

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

Solution steps

\frac{d}{d x} (4^{x} - 7)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (4^{x}) - \frac{d}{d x} (7)

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

\frac{d}{d x} (7) = 0

= ln (2) \cdot 2^{2 x + 1} - 0

Simplify

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

\frac{d y}{d x} + \frac{y}{x} = 3 x^{9} y^{2}

\int_{- 4}^{4} (1 + 16 - x^{2}) d x

\frac{d}{d x} (\frac{8 x}{( x ^{2} + 4 ) ^{2}})

\int \frac{x ^{5}}{x ^{6} + 7} d x

\frac{\partial}{\partial x} (\frac{4}{3} π r^{3})

What is the derivative of 4^x-7 ?
The derivative of 4^x-7 is ln(2)*2^{2x+1}
What is the first derivative of 4^x-7 ?
The first derivative of 4^x-7 is ln(2)*2^{2x+1}