What is the derivative of-2log_{e}(x)+x^2-4+e^x+x^e ?

The derivative of-2log_{e}(x)+x^2-4+e^x+x^e is 2x+ex^{e-1}+e^x-2/x

What is the first derivative of-2log_{e}(x)+x^2-4+e^x+x^e ?

The first derivative of-2log_{e}(x)+x^2-4+e^x+x^e is 2x+ex^{e-1}+e^x-2/x

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derivative of

- 2 lo g_{e} (x) + x^{2} - 4 + e^{x} + x^{e}

2 x + e x^{e - 1} + e^{x} - \frac{2}{x}

Solution steps

\frac{d}{d x} (- 2 lo g_{e} (x) + x^{2} - 4 + e^{x} + x^{e})

Simplify

- 2 lo g_{e} (x) + x^{2} - 4 + e^{x} + x^{e} : - 2 ln (x) + x^{2} - 4 + e^{x} + x^{e}

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (2 ln (x)) = \frac{2}{x}

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

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

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

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

= - \frac{2}{x} + 2 x - 0 + e^{x} + e x^{e - 1}

Simplify

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

\int 2 csc^{4} (x) cot^{6} (x) d x

x \to - 2 lim (7 - 2 x)

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

\frac{d}{d x} (\frac{2}{ln ( 10 ) x ^{3}})

\int_{4}^{7} \frac{1}{x ^{2} - 16} d x

What is the derivative of-2log_{e}(x)+x^2-4+e^x+x^e ?
The derivative of-2log_{e}(x)+x^2-4+e^x+x^e is 2x+ex^{e-1}+e^x-2/x
What is the first derivative of-2log_{e}(x)+x^2-4+e^x+x^e ?
The first derivative of-2log_{e}(x)+x^2-4+e^x+x^e is 2x+ex^{e-1}+e^x-2/x