What is the derivative of 6sqrt(xe^x) ?

The derivative of 6sqrt(xe^x) is (3(e^xx+e^x))/(sqrt(e^xx))

What is the first derivative of 6sqrt(xe^x) ?

The first derivative of 6sqrt(xe^x) is (3(e^xx+e^x))/(sqrt(e^xx))

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derivative of

6 x e^{x}

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

Solution steps

\frac{d}{d x} (6 x e^{x})

Take the constant out:

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

= 6 \frac{d}{d x} (x e^{x})

Apply the chain rule:

\frac{1}{2 x e ^{x}} \frac{d}{d x} (x e^{x})

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

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

= 6 \cdot \frac{1}{2 x e ^{x}} (e^{x} + e^{x} x)

Simplify

6 \cdot \frac{1}{2 x e ^{x}} (e^{x} + e^{x} x) : \frac{3 ( e ^{x} x + e ^{x} )}{e ^{x} x}

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

\frac{( 3 s + 2 )}{( s ^{2} + 4 s + 20 )}

x \to 0 + lim (\frac{x}{1 + ln ( x )})

y^{^{''}} + 7 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

\frac{d y}{d x} = 2 y x + y x^{2}, y (- 3) = 1

h (x) = \frac{1}{2 x}, at x = 4

What is the derivative of 6sqrt(xe^x) ?
The derivative of 6sqrt(xe^x) is (3(e^xx+e^x))/(sqrt(e^xx))
What is the first derivative of 6sqrt(xe^x) ?
The first derivative of 6sqrt(xe^x) is (3(e^xx+e^x))/(sqrt(e^xx))