What is the d/(dt)((x)(t^2+1)) ?

The d/(dt)((x)(t^2+1)) is (dx)/(dt)(t^2+1)+2tx

What is the first d/(dt)((x)(t^2+1)) ?

The first d/(dt)((x)(t^2+1)) is (dx)/(dt)(t^2+1)+2tx

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\frac{d}{d t} ((x) (t^{2} + 1))

\frac{d x}{d t} (t^{2} + 1) + 2 t x

Solution steps

\frac{d}{d t} ((x) (t^{2} + 1))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = t^{2} + 1

= \frac{d x}{d t} (t^{2} + 1) + \frac{d}{d t} (t^{2} + 1) x

\frac{d}{d t} (t^{2} + 1) = 2 t

= \frac{d x}{d t} (t^{2} + 1) + 2 t x

(x + 5)^{2}

(r sin (θ) - 1)^{2}

200 (- e^{\frac{1}{2}} + 1)

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

(\frac{3}{5} x - \frac{4}{5} y)^{2}

What is the d/(dt)((x)(t^2+1)) ?
The d/(dt)((x)(t^2+1)) is (dx)/(dt)(t^2+1)+2tx
What is the first d/(dt)((x)(t^2+1)) ?
The first d/(dt)((x)(t^2+1)) is (dx)/(dt)(t^2+1)+2tx