What is the d/(dt)((t)(k(M-t))) ?

The d/(dt)((t)(k(M-t))) is k(M-2t)

What is the first d/(dt)((t)(k(M-t))) ?

The first d/(dt)((t)(k(M-t))) is k(M-2t)

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\frac{d}{d t} ((t) (k (M - t)))

k (M - 2 t)

Solution steps

\frac{d}{d t} ((t) (k (M - t)))

Treat

k, M

as a constant

Take the constant out:

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

= k \frac{d}{d t} (t (M - t))

Apply the Product Rule:

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

f = t, g = M - t

= k (\frac{d t}{d t} (M - t) + \frac{d}{d t} (M - t) t)

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

\frac{d}{d t} (M - t) = - 1

= k (1 \cdot (M - t) + (- 1) t)

Simplify

k (1 \cdot (M - t) + (- 1) t) : k (M - 2 t)

= k (M - 2 t)

\frac{2 x ^{5} + x ^{4} + 4 x ^{3} + 3 x ^{2} - 24 x}{( x ^{2} + 1 ) ^{2}}

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\frac{3 x ^{2} ( 3 x ^{2} + 2 x ^{4} )}{8}

\frac{( s + 2 )}{( s ^{2} + 4 s + 3 ) \cdot ( s + 5 )}

\frac{4 s + 1}{5 ( s ^{2} + 2 s + 2 )}

What is the d/(dt)((t)(k(M-t))) ?
The d/(dt)((t)(k(M-t))) is k(M-2t)
What is the first d/(dt)((t)(k(M-t))) ?
The first d/(dt)((t)(k(M-t))) is k(M-2t)