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

The derivative of 3e^{x^2+x} is 3e^{x^2+x}(2x+1)

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

The first derivative of 3e^{x^2+x} is 3e^{x^2+x}(2x+1)

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

3 e^{x^{2} + x} (2 x + 1)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

e^{x^{2} + x} \frac{d}{d x} (x^{2} + x)

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

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

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

n = 1 \sum \infty \frac{e ^{n}}{π ^{n}}

\frac{d}{d x} (\frac{5 ^{x^{3}}}{4 x ^{2}})

y y^{^{'}} = \frac{7}{x}

(1 + e^{2 x}) y^{2} d y = e^{x} d x, y (0) = 0

\int 2 x - \frac{1}{1 + x ^{2}} d x

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