What is the derivative of f(x)=(x^2+1)*e^{x-1} ?

The derivative of f(x)=(x^2+1)*e^{x-1} is 2xe^{x-1}+e^{x-1}(x^2+1)

What is the first derivative of f(x)=(x^2+1)*e^{x-1} ?

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

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derivative of

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

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