What is the d/(du)((x(u,t))(t^2+5t)) ?

The d/(du)((x(u,t))(t^2+5t)) is x(t^2+5t)(,t*1)

What is the first d/(du)((x(u,t))(t^2+5t)) ?

The first d/(du)((x(u,t))(t^2+5t)) is x(t^2+5t)(,t*1)

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\frac{d}{d u} ((x (u, t)) (t^{2} + 5 t))

x (t^{2} + 5 t) (, t \cdot 1)

Solution steps

\frac{d}{d u} ((x (u, t)) (t^{2} + 5 t))

Treat

x, t

as constants

Take the constant out:

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

= x (t^{2} + 5 t) \frac{d}{d u} (u, t)

Take the constant out:

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

= x (t^{2} + 5 t) (, t \frac{d u}{d u})

Apply the common derivative:

\frac{d u}{d u} = 1

= x (t^{2} + 5 t) (, t \cdot 1)

(ln (x) - 1)^{3}

\frac{d ^{2}}{d x ^{2}} ((y) (6 x - 3))

\frac{7 s - 16}{( s - 2 ) ^{2} + 16}

∣ A - B ∣^{2}

\frac{4}{( 2 x ^{8} - 5 ) ^{4}}

What is the d/(du)((x(u,t))(t^2+5t)) ?
The d/(du)((x(u,t))(t^2+5t)) is x(t^2+5t)(,t*1)
What is the first d/(du)((x(u,t))(t^2+5t)) ?
The first d/(du)((x(u,t))(t^2+5t)) is x(t^2+5t)(,t*1)