What is the derivative of 1+3^{2x} ?

The derivative of 1+3^{2x} is 2ln(3)*9^x

What is the first derivative of 1+3^{2x} ?

The first derivative of 1+3^{2x} is 2ln(3)*9^x

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

2 ln (3) \cdot 9^{x}

Solution steps

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

Simplify

1 + 3^{2 x} : 1 + 9^{x}

= \frac{d}{d x} (1 + 9^{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (1) + \frac{d}{d x} (9^{x})

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

\frac{d}{d x} (9^{x}) = 2 ln (3) \cdot 9^{x}

= 0 + 2 ln (3) \cdot 9^{x}

Simplify

= 2 ln (3) \cdot 9^{x}

\int \frac{4}{6 + 4 x} d x

\int_{- r}^{r} (r^{2} - x^{2}) d x

y^{^{'}} = 0.21 - \frac{y}{1700}

\int_{- 5}^{0} x 4 - x d x

f (x) = sin (2 x) cos (x)

What is the derivative of 1+3^{2x} ?
The derivative of 1+3^{2x} is 2ln(3)*9^x
What is the first derivative of 1+3^{2x} ?
The first derivative of 1+3^{2x} is 2ln(3)*9^x