What is the derivative of 4^x+1 ?

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

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

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

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derivative of

4^{x} + 1

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (1) = 0

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

Simplify

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

\frac{d}{d a} (\frac{a ^{2} - 2 a + a}{9})

y^{2} x + 5 x y^{3} + 10 y^{2} + 8 x y

\frac{d}{d a} (\frac{( cos ( a ) )}{( 1 - sin ( a ) )})

y = \frac{1}{( x ^{2} + 4 )}

(1 + 3 x)^{\frac{1}{3}}

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