What is the derivative of mxe^{nx^2} ?

The derivative of mxe^{nx^2} is m(e^{nx^2}+2ne^{nx^2}x^2)

What is the first derivative of mxe^{nx^2} ?

The first derivative of mxe^{nx^2} is m(e^{nx^2}+2ne^{nx^2}x^2)

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\frac{d}{d x} (m x e^{n x^{2}})

m (e^{n x^{2}} + 2 n e^{n x^{2}} x^{2})

Solution steps

\frac{d}{d x} (m x e^{n x^{2}})

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = e^{n x^{2}}

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

\frac{d x}{d x} = 1

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

= m (1 \cdot e^{n x^{2}} + e^{n x^{2}} \cdot 2 n xx)

Simplify

m (1 \cdot e^{n x^{2}} + e^{n x^{2}} \cdot 2 n xx) : m (e^{n x^{2}} + 2 n e^{n x^{2}} x^{2})

= m (e^{n x^{2}} + 2 n e^{n x^{2}} x^{2})

\frac{d}{d x} (5 \cdot T g (T, x) (2 x^{3}) + \frac{2}{x ^{2}})

y = \frac{5}{x}, (1, 5)

\int_{\frac{6}{5}}^{4} \frac{1}{1 - x ^{2}} d x

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

x \to \infty lim (0.72 x)

What is the derivative of mxe^{nx^2} ?
The derivative of mxe^{nx^2} is m(e^{nx^2}+2ne^{nx^2}x^2)
What is the first derivative of mxe^{nx^2} ?
The first derivative of mxe^{nx^2} is m(e^{nx^2}+2ne^{nx^2}x^2)