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

The derivative of (x^2-1)*e^x is 2xe^x+e^x(x^2-1)

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

The first derivative of (x^2-1)*e^x is 2xe^x+e^x(x^2-1)

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derivative of

(x^{2} - 1) \cdot e^{x}

2 x e^{x} + e^{x} (x^{2} - 1)

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

f (x) = \frac{7}{x}, at x = 5

y^{^{'}} + e^{x} y = 0

\int x - 6 d x

x \to 0 lim (\frac{[ sin ( 8 x ) ]}{x})

y^{^{''}} + y = 0, y (\frac{π}{3}) = 12, y^{^{'}} (\frac{π}{3}) = - 24

What is the derivative of (x^2-1)*e^x ?
The derivative of (x^2-1)*e^x is 2xe^x+e^x(x^2-1)
What is the first derivative of (x^2-1)*e^x ?
The first derivative of (x^2-1)*e^x is 2xe^x+e^x(x^2-1)