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

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

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

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

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\frac{d}{d x} ((x^{3} - 4 x + 1) e^{x})

e^{x} x^{3} + 3 e^{x} x^{2} - 4 e^{x} x - 3 e^{x}

Solution steps

\frac{d}{d x} ((x^{3} - 4 x + 1) e^{x})

Apply the Product Rule:

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

f = x^{3} - 4 x + 1, g = e^{x}

= \frac{d}{d x} (x^{3} - 4 x + 1) e^{x} + \frac{d}{d x} (e^{x}) (x^{3} - 4 x + 1)

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

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

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

Simplify

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

= e^{x} x^{3} + 3 e^{x} x^{2} - 4 e^{x} x - 3 e^{x}

x \to 0 lim (\frac{5}{tan ( 3 x )})

8 \frac{d y}{d x} + 7 e^{x + y} = 0

y^{^{'}} = t y

(t^{2} + 1) f^{^{'}} (t) + [f (t)]^{2} + 1 = 0

\frac{d}{d a} (cos (3 a) cos (2 a) + sin (4 a) sin (a))

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