What is the derivative of 3^x-2^x ?

The derivative of 3^x-2^x is 3^xln(3)-2^xln(2)

What is the first derivative of 3^x-2^x ?

The first derivative of 3^x-2^x is 3^xln(3)-2^xln(2)

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

3^{x} ln (3) - 2^{x} ln (2)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (3^{x}) = 3^{x} ln (3)

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

= 3^{x} ln (3) - 2^{x} ln (2)

\frac{d}{d x} (x 6 - x^{2})

\int - 2 sin (t) d t

n = 2 \sum \infty \frac{7}{2 ^{n}}

\int_{0}^{1} \frac{32}{x ^{2} - 64} d x

\frac{1}{s ^{2} \cdot ( s + 1 )}

What is the derivative of 3^x-2^x ?
The derivative of 3^x-2^x is 3^xln(3)-2^xln(2)
What is the first derivative of 3^x-2^x ?
The first derivative of 3^x-2^x is 3^xln(3)-2^xln(2)