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)

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)