What is the derivative of 38xe^{x^2} ?

The derivative of 38xe^{x^2} is 38(e^{x^2}+2e^{x^2}x^2)

What is the first derivative of 38xe^{x^2} ?

The first derivative of 38xe^{x^2} is 38(e^{x^2}+2e^{x^2}x^2)

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derivative of

38 x e^{x^{2}}

38 (e^{x^{2}} + 2 e^{x^{2}} x^{2})

Solution steps

\frac{d}{d x} (38 x e^{x^{2}})

Take the constant out:

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

= 38 \frac{d}{d x} (x e^{x^{2}})

Apply the Product Rule:

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

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

= 38 (\frac{d x}{d x} e^{x^{2}} + \frac{d}{d x} (e^{x^{2}}) x)

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

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

= 38 (1 \cdot e^{x^{2}} + e^{x^{2}} \cdot 2 xx)

Simplify

38 (1 \cdot e^{x^{2}} + e^{x^{2}} \cdot 2 xx) : 38 (e^{x^{2}} + 2 e^{x^{2}} x^{2})

= 38 (e^{x^{2}} + 2 e^{x^{2}} x^{2})

\frac{\partial}{\partial y} (5 - 2 x^{2} - y^{2})

y^{^{''}} + 4 y = 16 sec (2 t)

\frac{d}{d t} (4 t^{3})

\int_{0}^{\frac{π}{2}} cos^{5} (x) d x

2 x^{4} + \frac{3 x}{5}, at x = 0

What is the derivative of 38xe^{x^2} ?
The derivative of 38xe^{x^2} is 38(e^{x^2}+2e^{x^2}x^2)
What is the first derivative of 38xe^{x^2} ?
The first derivative of 38xe^{x^2} is 38(e^{x^2}+2e^{x^2}x^2)