What is expand-(t(1+in))/(dt) ?

The solution to expand-(t(1+in))/(dt) is -t/(dt)-(int)/(dt)

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expand-(t(1+in))/(dt)

expand

- \frac{t ( 1 + in )}{d t}

- \frac{t}{d t} - \frac{in t}{d t}

Solution steps

- \frac{t ( 1 + in )}{d t}

Expand

t (1 + in) : t + in t

= - \frac{t + in t}{d t}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{t + in t}{d t} = - (\frac{t}{d t}) - (\frac{in t}{d t})

= - (\frac{t}{d t}) - (\frac{in t}{d t})

Remove parentheses:

(a) = a

= - \frac{t}{d t} - \frac{in t}{d t}

(1 - \frac{x}{a})^{2}

\frac{a ^{2} - 1}{a} - \frac{1}{a - 1}

- \frac{d}{d s} (\frac{e ^{- π s}}{s ( s ^{2} + 1 )})

\frac{9 x ^{3} + 16 x ^{2} + 3 x - 10}{x ^{3} ( x + 5 )}

(x + 1 + 2) (x + 1 - 2) (x^{2} + 7)

What is expand-(t(1+in))/(dt) ?
The solution to expand-(t(1+in))/(dt) is -t/(dt)-(int)/(dt)