What is (\partial)/(\partial x)(x*e^{x^2-4}) ?

The answer to (\partial)/(\partial x)(x*e^{x^2-4}) is e^{x^2-4}+2e^{x^2-4}x^2

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(\partial)/(\partial x)(x*e^{x^2-4})

\frac{\partial}{\partial x} (x \cdot e^{x^{2} - 4})

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

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (e^{x^{2} - 4}) = e^{x^{2} - 4} \cdot 2 x

= 1 \cdot e^{x^{2} - 4} + e^{x^{2} - 4} \cdot 2 xx

Simplify

1 \cdot e^{x^{2} - 4} + e^{x^{2} - 4} \cdot 2 xx : e^{x^{2} - 4} + 2 e^{x^{2} - 4} x^{2}

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

\frac{d}{d x} ((ln (6 x))^{2})

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

f (t) = \frac{2 t}{2 + t}

- cos (t)

x^{3} - x

What is (\partial)/(\partial x)(x*e^{x^2-4}) ?
The answer to (\partial)/(\partial x)(x*e^{x^2-4}) is e^{x^2-4}+2e^{x^2-4}x^2