What is the derivative of-2ln(x+x^2-4+e^x+x^e) ?

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

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

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

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\frac{d}{d x} (- 2 ln (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 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}

\frac{\partial}{\partial x} (2 yz)

x \to 6 - lim (e^{\frac{3}{6 - x}})

\frac{d y}{d x} = y x

\frac{d}{d x} ((10 - 5 x^{2})^{\frac{4}{3}})

f (x) = 7 x - 7 x

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