The anti derivative of (ax + b)2 is a function of x whose derivative is (ax + b)2.
It is known that,
\begin{align} \frac {d}{dx} ((ax+b)^3) = 3a(ax+b)^2 \end{align}
⇒ \begin{align} (ax + b)^2 =\frac {1}{3a} \frac {d}{dx}(ax+b)^3 \end{align}
∴ \begin{align} (ax + b)^2 = \frac {d}{dx}\left(\frac {1}{3a}(ax + b)^3\right) \end{align}
Therefore, the anti derivative of (ax +b)2
\begin{align} (ax + B)^2 \;is \frac {1}{3a}(ax +b)^3 \end{align}
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Subject to the constraints:x + y ≤ 4, x ≥ 0, y ≥ 0
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\begin{align}\left(\frac{d^2y}{dx^2}\right)^3\;+ \left(\frac{dy}{dx}\right)^2+\;sin\left(\frac{dy}{dx}\right)\;+ 1=\;0\end{align}
is (A) 3 (B) 2 (C) 1 (D) not defined
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R (x) = 13x2 + 26x + 15
Find the marginal revenue when x = 7.
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(f + g)oh = foh + goh
(f . g)oh = (foh) . (goh)
Maximise Z = 3x + 4y
Subject to the constraints:x + y ≤ 4, x ≥ 0, y ≥ 0