What is the d/(da)(abcd+7ab-2bc+5cd-5) ?

The d/(da)(abcd+7ab-2bc+5cd-5) is bcd+7b

What is the first d/(da)(abcd+7ab-2bc+5cd-5) ?

The first d/(da)(abcd+7ab-2bc+5cd-5) is bcd+7b

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\frac{d}{d a} (ab c d + 7 ab - 2 b c + 5 c d - 5)

b c d + 7 b

Solution steps

\frac{d}{d a} (ab c d + 7 ab - 2 b c + 5 c d - 5)

Apply the Sum/Difference Rule:

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

= \frac{d}{d a} (ab c d) + \frac{d}{d a} (7 ab) - \frac{d}{d a} (2 b c) + \frac{d}{d a} (5 c d) - \frac{d}{d a} (5)

\frac{d}{d a} (ab c d) = b c d

\frac{d}{d a} (7 ab) = 7 b

\frac{d}{d a} (2 b c) = 0

\frac{d}{d a} (5 c d) = 0

\frac{d}{d a} (5) = 0

= b c d + 7 b - 0 + 0 - 0

Simplify

= b c d + 7 b

\frac{d y}{d x} + 1 = 2 y, y (1) = 1

\int \frac{2 u - 1}{4 u ^{2} - 1} d u

\int \frac{4 y}{1} d y

y = 2 e^{x} cos (x), (0, 2)

\int x^{π} d x

What is the d/(da)(abcd+7ab-2bc+5cd-5) ?
The d/(da)(abcd+7ab-2bc+5cd-5) is bcd+7b
What is the first d/(da)(abcd+7ab-2bc+5cd-5) ?
The first d/(da)(abcd+7ab-2bc+5cd-5) is bcd+7b