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

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

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

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

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

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

Solution steps

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

Apply the Product Rule:

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

f = x, g = (t - 1)^{2}

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

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

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

lo g_{\frac{x ^{2} 4 - x}{( x - 1 ) ^{4} ( 2 x - 1 ) ^{3}}} (x)

(2 i - 3)^{4}

(sec (x) + tan (x))^{3}

\frac{n ( 3 + n )}{18}

- 4 (2 x e^{2 x} + e^{2 x} \cdot 2 x^{2})

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