What is the derivative of (8x^2-14xe^x) ?

The derivative of (8x^2-14xe^x) is 8e^xx^2+2e^xx-14e^x

What is the first derivative of (8x^2-14xe^x) ?

The first derivative of (8x^2-14xe^x) is 8e^xx^2+2e^xx-14e^x

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\frac{d}{d x} ((8 x^{2} - 14 x) e^{x})

8 e^{x} x^{2} + 2 e^{x} x - 14 e^{x}

Solution steps

\frac{d}{d x} ((8 x^{2} - 14 x) e^{x})

Apply the Product Rule:

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

f = 8 x^{2} - 14 x, g = e^{x}

= \frac{d}{d x} (8 x^{2} - 14 x) e^{x} + \frac{d}{d x} (e^{x}) (8 x^{2} - 14 x)

\frac{d}{d x} (8 x^{2} - 14 x) = 16 x - 14

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

= (16 x - 14) e^{x} + e^{x} (8 x^{2} - 14 x)

Simplify

(16 x - 14) e^{x} + e^{x} (8 x^{2} - 14 x) : 8 e^{x} x^{2} + 2 e^{x} x - 14 e^{x}

= 8 e^{x} x^{2} + 2 e^{x} x - 14 e^{x}

\int_{1}^{25} e^{x} d x

\frac{d}{d x} (ln ((x + 3)^{4}))

x \to \frac{1}{2} lim (\frac{1}{( 2 x - 1 )})

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

y^{^{''}} + 4 y^{^{'}} - 12 y = 0

What is the derivative of (8x^2-14xe^x) ?
The derivative of (8x^2-14xe^x) is 8e^xx^2+2e^xx-14e^x
What is the first derivative of (8x^2-14xe^x) ?
The first derivative of (8x^2-14xe^x) is 8e^xx^2+2e^xx-14e^x