What is the derivative of 6*3^x ?

The derivative of 6*3^x is 2ln(3)*3^{1+x}

What is the first derivative of 6*3^x ?

The first derivative of 6*3^x is 2ln(3)*3^{1+x}

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\frac{d}{d x} (6 \cdot 3^{x})

2 ln (3) \cdot 3^{1 + x}

Solution steps

\frac{d}{d x} (6 \cdot 3^{x})

Simplify

6 \cdot 3^{x} : 2 \cdot 3^{1 + x}

= \frac{d}{d x} (2 \cdot 3^{1 + x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (3^{1 + x})

Apply exponent rule:

a^{b} = e^{b l n (a)}

3^{1 + x} = e^{(1 + x) l n (3)}

= 2 \frac{d}{d x} (e^{(1 + x) l n (3)})

Apply the chain rule:

e^{(1 + x) l n (3)} \frac{d}{d x} ((1 + x) ln (3))

= e^{(1 + x) l n (3)} \frac{d}{d x} ((1 + x) ln (3))

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

= 2 e^{(1 + x) l n (3)} ln (3)

e^{(1 + x) l n (3)} = 3^{1 + x}

= 2 ln (3) \cdot 3^{1 + x}

y = 2 - \frac{x}{2}, x = 2 y^{2}

\int (9 x + 2)^{2} d x

\int \frac{a + b x ^{5}}{6 a x + b x ^{6}} d x

y^{^{''}} + 4 y^{^{'}} - 24 y = 0

\int e^{x} + e^{x} x d x

What is the derivative of 6*3^x ?
The derivative of 6*3^x is 2ln(3)*3^{1+x}
What is the first derivative of 6*3^x ?
The first derivative of 6*3^x is 2ln(3)*3^{1+x}