What is the derivative of e^x(2x^2+x-8) ?

The derivative of e^x(2x^2+x-8) is 2e^xx^2+5e^xx-7e^x

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

The first derivative of e^x(2x^2+x-8) is 2e^xx^2+5e^xx-7e^x

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

2 e^{x} x^{2} + 5 e^{x} x - 7 e^{x}

Solution steps

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

Apply the Product Rule:

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

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

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

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

\frac{d}{d x} (2 x^{2} + x - 8) = 4 x + 1

= e^{x} (2 x^{2} + x - 8) + (4 x + 1) e^{x}

Simplify

e^{x} (2 x^{2} + x - 8) + (4 x + 1) e^{x} : 2 e^{x} x^{2} + 5 e^{x} x - 7 e^{x}

= 2 e^{x} x^{2} + 5 e^{x} x - 7 e^{x}

\int - 4 x^{4} + 8 x^{3} d x

\int \frac{2 x - 3}{x ^{2} + 2 x - 17} d x

y^{\frac{1}{2}} \frac{d y}{d x} + y^{\frac{3}{2}} = 1, y (0) = 16

\int \frac{sin ( 2 θ )}{1 + sin ^{2} ( θ )} d θ

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

What is the derivative of e^x(2x^2+x-8) ?
The derivative of e^x(2x^2+x-8) is 2e^xx^2+5e^xx-7e^x
What is the first derivative of e^x(2x^2+x-8) ?
The first derivative of e^x(2x^2+x-8) is 2e^xx^2+5e^xx-7e^x