What is expand-4(x+1/(e^x)-1/(2e^{2x)}-ln|e^x+1|) ?

The solution to expand-4(x+1/(e^x)-1/(2e^{2x)}-ln|e^x+1|) is -4x-4/(e^x)+2/(e^{2x)}+4ln|e^x+1|

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expand-4(x+1/(e^x)-1/(2e^{2x)}-ln|e^x+1|)

expand

- 4 (x + \frac{1}{e ^{x}} - \frac{1}{2 e ^{2 x}} - ln ∣ e^{x} + 1 ∣)

- 4 x - \frac{4}{e ^{x}} + \frac{2}{e ^{2 x}} + 4 ln ∣ e^{x} + 1 ∣

Solution steps

- 4 (x + \frac{1}{e ^{x}} - \frac{1}{2 e ^{2 x}} - ln ∣ e^{x} + 1 ∣)

Distribute parentheses

= (- 4) x + (- 4) \frac{1}{e ^{x}} + (- 4) (- \frac{1}{2 e ^{2 x}}) + (- 4) (- ln ∣ e^{x} + 1 ∣)

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= - 4 x - 4 \cdot \frac{1}{e ^{x}} + 4 \cdot \frac{1}{2 e ^{2 x}} + 4 ln ∣ e^{x} + 1 ∣

Simplify

- 4 x - 4 \cdot \frac{1}{e ^{x}} + 4 \cdot \frac{1}{2 e ^{2 x}} + 4 ln ∣ e^{x} + 1 ∣ : - 4 x - \frac{4}{e ^{x}} + \frac{2}{e ^{2 x}} + 4 ln ∣ e^{x} + 1 ∣

= - 4 x - \frac{4}{e ^{x}} + \frac{2}{e ^{2 x}} + 4 ln ∣ e^{x} + 1 ∣

\frac{x ^{n + 1}}{n ! ( n + 1 )}

\frac{\partial}{\partial x} ((w) (u (w, v, x)^{2} + e^{2 v}))

2 sec^{2} (x) tan^{2} (x) (tan (x) + sec (x))

\frac{x ^{n + 1}}{( n + 1 ) n !}

\frac{( 18 n - 7 ) ^{2}}{441}

What is expand-4(x+1/(e^x)-1/(2e^{2x)}-ln|e^x+1|) ?
The solution to expand-4(x+1/(e^x)-1/(2e^{2x)}-ln|e^x+1|) is -4x-4/(e^x)+2/(e^{2x)}+4ln|e^x+1|