What is the (e^{2x}-x^2-1)^' ?

The (e^{2x}-x^2-1)^' is -2x+2e^{2x}

What is the first (e^{2x}-x^2-1)^' ?

The first (e^{2x}-x^2-1)^' is -2x+2e^{2x}

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

- 2 x + 2 e^{2 x}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

(e^{2 x})^{'} = e^{2 x} \cdot 2

(x^{2})^{'} = 2 x

1^{'} = 0

= e^{2 x} \cdot 2 - 2 x - 0

Simplify

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

\int_{- 1}^{2} x^{2} + 1 d x

\int u 1 - u^{2} d u

\int x \cdot e^{- 2 x} d x

\frac{d}{d x} (\frac{1}{10 x})

14 - x^{2}, x^{2} - 4