What is the derivative of y=3^x+1 ?

The derivative of y=3^x+1 is ln(3)*3^x

What is the first derivative of y=3^x+1 ?

The first derivative of y=3^x+1 is ln(3)*3^x

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derivative of

y = 3^{x} + 1

ln (3) \cdot 3^{x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

\int (sin (x) - cos (2 x))^{2} d x

\int (7 + 7 tan^{2} (θ)) d θ

\frac{d}{d x} (\frac{x ^{4} - 3 x ^{3} - 6 x ^{2} - 3 x + 2}{x})

f (x) = \frac{x}{( 2 x - 1 ) ^{6}}, at x = 1

\frac{d y}{d x} + 3 y + 5 = 0

What is the derivative of y=3^x+1 ?
The derivative of y=3^x+1 is ln(3)*3^x
What is the first derivative of y=3^x+1 ?
The first derivative of y=3^x+1 is ln(3)*3^x