What is the derivative of e^x(2x^4-1) ?

The derivative of e^x(2x^4-1) is e^x(2x^4-1)+8x^3e^x

What is the first derivative of e^x(2x^4-1) ?

The first derivative of e^x(2x^4-1) is e^x(2x^4-1)+8x^3e^x

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\frac{d}{d x} (e^{x} (2 x^{4} - 1))

e^{x} (2 x^{4} - 1) + 8 x^{3} e^{x}

Solution steps

\frac{d}{d x} (e^{x} (2 x^{4} - 1))

Apply the Product Rule:

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

f = e^{x}, g = 2 x^{4} - 1

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

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

\frac{d}{d x} (2 x^{4} - 1) = 8 x^{3}

= e^{x} (2 x^{4} - 1) + 8 x^{3} e^{x}

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

\int x \cdot ln (x) d x

\int_{- 4}^{0} (1 + 16 - x^{2}) d x

\int t^{2} 13 t - 1 d t

f (x) = (\frac{4}{x}) + (\frac{1}{5 x ^{3} + 2 x})

What is the derivative of e^x(2x^4-1) ?
The derivative of e^x(2x^4-1) is e^x(2x^4-1)+8x^3e^x
What is the first derivative of e^x(2x^4-1) ?
The first derivative of e^x(2x^4-1) is e^x(2x^4-1)+8x^3e^x