What is the d/(dt)((x)(bt+ct^2)) ?

The d/(dt)((x)(bt+ct^2)) is x(b+2ct)

What is the first d/(dt)((x)(bt+ct^2)) ?

The first d/(dt)((x)(bt+ct^2)) is x(b+2ct)

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\frac{d}{d t} ((x) (b t + c t^{2}))

x (b + 2 c t)

Solution steps

\frac{d}{d t} ((x) (b t + c t^{2}))

Treat

b, c, x

as a constant

Take the constant out:

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

= x \frac{d}{d t} (b t + c t^{2})

Apply the Sum/Difference Rule:

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

= x (\frac{d}{d t} (b t) + \frac{d}{d t} (c t^{2}))

\frac{d}{d t} (b t) = b

\frac{d}{d t} (c t^{2}) = 2 c t

= x (b + 2 c t)

\frac{( 2 x + x ^{3} ) ^{3}}{12}

(y^{2} + 3 x) d x + (e^{x}) d y

(1, i) (2 - 1, 3 + 2 i)

\frac{( x + 3 ) ( 3 x + 1 + 1 )}{3}

e^{x} (x^{3} - 3 x)

What is the d/(dt)((x)(bt+ct^2)) ?
The d/(dt)((x)(bt+ct^2)) is x(b+2ct)
What is the first d/(dt)((x)(bt+ct^2)) ?
The first d/(dt)((x)(bt+ct^2)) is x(b+2ct)