What is the derivative of 3^x-4 ?

The derivative of 3^x-4 is ln(3)*3^x

What is the first derivative of 3^x-4 ?

The first derivative of 3^x-4 is ln(3)*3^x

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

ln (3) \cdot 3^{x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (4) = 0

= 3^{x} ln (3) - 0

Simplify

= ln (3) \cdot 3^{x}

\frac{\partial}{\partial x} (e^{x y} + x y e^{x y})

\frac{d}{d x} (\frac{a x ^{3} + b x + c}{d x ^{3} - e})

\int_{0}^{1} \frac{6 + x}{1 + x ^{2}} d x

f (x) = \frac{x ^{3}}{x + 1}, at x = 1

\frac{d}{d x} (tan (π x - 1) + e^{c o s (x)})

What is the derivative of 3^x-4 ?
The derivative of 3^x-4 is ln(3)*3^x
What is the first derivative of 3^x-4 ?
The first derivative of 3^x-4 is ln(3)*3^x