What is the derivative of 4^x-2^x ?

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

What is the first derivative of 4^x-2^x ?

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

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derivative of

4^{x} - 2^{x}

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

Solution steps

\frac{d}{d x} (4^{x} - 2^{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (4^{x}) - \frac{d}{d x} (2^{x})

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

\frac{d}{d x} (2^{x}) = 2^{x} ln (2)

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

\frac{\partial}{\partial x} (2 x^{2} + 9 y^{2} + 3)

\frac{d}{d x} (2 x^{3} + 4 x - 2)

4 x^{3} - 16 x

\frac{d x}{d t} - 4 x = cos (4 t)

\int 8 sin (- 2 t + 7) d t

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