What is the derivative of f(x)=(8x^2-14x)e^x ?

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

What is the first derivative of f(x)=(8x^2-14x)e^x ?

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

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derivative of

f (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}

\frac{3}{( s + 1 )}

x \to - 1 lim (5^{x^{2} - 3 x})

\int \frac{x ^{3}}{( x ^{4} + 3 ) ^{\frac{7}{2}}} d x

\int \frac{e ^{x}}{e ^{2 x} + 4} d x

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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