What is the d/(dt)((x)(-x-y(x,t)-t)) ?

The d/(dt)((x)(-x-y(x,t)-t)) is x(-0-y(x,1)-1)

What is the first d/(dt)((x)(-x-y(x,t)-t)) ?

The first d/(dt)((x)(-x-y(x,t)-t)) is x(-0-y(x,1)-1)

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\frac{d}{d t} ((x) (- x - y (x, t) - t))

x (- 0 - y (x, 1) - 1)

Solution steps

\frac{d}{d t} ((x) (- x - y (x, t) - t))

Treat

x, y

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= x \frac{d}{d t} (- x - y (x, t) - t)

Apply the Sum/Difference Rule:

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

= x (- \frac{d x}{d t} - \frac{d}{d t} (y (x, t)) - \frac{d t}{d t})

\frac{d x}{d t} = 0

\frac{d}{d t} (y (x, t)) = y (x, 1)

\frac{d t}{d t} = 1

= x (- 0 - y (x, 1) - 1)

\frac{2 x - 1}{( a x + b ) ( c x + d ) ^{2}}

\frac{( ( x + h ) ^{\frac{2}{3}} - x ^{\frac{2}{3}} )}{h}

\frac{2 ( x + 9 ) ( x - 1 )}{( x + 4 ) ( x - 2 )}

\frac{1}{( x + 1 ) ( x - 4 )}

- (u - z u) (z d u + u d z)

What is the d/(dt)((x)(-x-y(x,t)-t)) ?
The d/(dt)((x)(-x-y(x,t)-t)) is x(-0-y(x,1)-1)
What is the first d/(dt)((x)(-x-y(x,t)-t)) ?
The first d/(dt)((x)(-x-y(x,t)-t)) is x(-0-y(x,1)-1)