What is the -derivative of e^{3x} ?

The -derivative of e^{3x} is -x derivative of e^{3x}+3e^{3x}x-e^{3x}+C

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Find the Derivative - d/dx e^(3x)

\int - \frac{d}{d x} (e^{3 x}) d x

- x \frac{d}{d x} (e^{3 x}) + 3 e^{3 x} x - e^{3 x} + C

Solution steps

\int - \frac{d}{d x} (e^{3 x}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

: \frac{d}{d x} (e^{3 x}) x - \int 9 e^{3 x} x d x

= - (\frac{d}{d x} (e^{3 x}) x - \int 9 e^{3 x} x d x)

\int 9 e^{3 x} x d x = 3 e^{3 x} x - e^{3 x}

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

Simplify

- (\frac{d}{d x} (e^{3 x}) x - (3 e^{3 x} x - e^{3 x})) : - x \frac{d}{d x} (e^{3 x}) + 3 e^{3 x} x - e^{3 x}

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

Add a constant to the solution

= - x \frac{d}{d x} (e^{3 x}) + 3 e^{3 x} x - e^{3 x} + C

(4, 0), (0, 3)

x \to \infty lim (\frac{x}{4})

\frac{d y}{d x} y = cos (x)

- \frac{d}{d x} (cos (3 x))

f (x) = arctan (x)

What is the -derivative of e^{3x} ?
The -derivative of e^{3x} is -x derivative of e^{3x}+3e^{3x}x-e^{3x}+C