What is the derivative of g(0)=(4-2x)e^n-4 ?

The derivative of g(0)=(4-2x)e^n-4 is -2e^n

What is the first derivative of g(0)=(4-2x)e^n-4 ?

The first derivative of g(0)=(4-2x)e^n-4 is -2e^n

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derivative of

g (0) = (4 - 2 x) e^{n} - 4

- 2 e^{n}

Solution steps

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

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} ((4 - 2 x) e^{n}) - \frac{d}{d x} (4)

\frac{d}{d x} ((4 - 2 x) e^{n}) = - 2 e^{n}

\frac{d}{d x} (4) = 0

= - 2 e^{n} - 0

Simplify

= - 2 e^{n}

x \to - 1 lim (5 - x^{2})

\frac{π x}{4}

y^{^{'}} = 4 x (x - 2)

(1 + x)^{\frac{3}{2}}

\frac{d}{d x} (1 0^{1 - x^{2}})

What is the derivative of g(0)=(4-2x)e^n-4 ?
The derivative of g(0)=(4-2x)e^n-4 is -2e^n
What is the first derivative of g(0)=(4-2x)e^n-4 ?
The first derivative of g(0)=(4-2x)e^n-4 is -2e^n