What is the d/(dt)((x)(100(1-x^2))) ?

The d/(dt)((x)(100(1-x^2))) is 100(1-3t^2)

What is the first d/(dt)((x)(100(1-x^2))) ?

The first d/(dt)((x)(100(1-x^2))) is 100(1-3t^2)

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\frac{d}{d t} ((t) (100 (1 - t^{2})))

100 (1 - 3 t^{2})

Solution steps

\frac{d}{d t} ((t) (100 (1 - t^{2})))

Take the constant out:

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

= 100 \frac{d}{d t} (t (1 - t^{2}))

Apply the Product Rule:

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

f = t, g = 1 - t^{2}

= 100 (\frac{d t}{d t} (1 - t^{2}) + \frac{d}{d t} (1 - t^{2}) t)

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

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

= 100 (1 \cdot (1 - t^{2}) + (- 2 t) t)

Simplify

100 (1 \cdot (1 - t^{2}) + (- 2 t) t) : 100 (1 - 3 t^{2})

= 100 (1 - 3 t^{2})

\frac{1}{1 + x}

\frac{2}{( s ^{2} + 1 ) ( s - 1 ) ^{2}}

\frac{( 100 + x ) ( 56 + 3 x )}{35}

\frac{[ 5 ( x - h ) ] - [ 5 x ^{2} ]}{h}

(2^{6})^{x + 3}

What is the d/(dt)((x)(100(1-x^2))) ?
The d/(dt)((x)(100(1-x^2))) is 100(1-3t^2)
What is the first d/(dt)((x)(100(1-x^2))) ?
The first d/(dt)((x)(100(1-x^2))) is 100(1-3t^2)