Differentiate the function y sqrt 3 x 2 2x - x 2 - JAMB Mathematics 2023 Question
Differentiate the function y = \(\sqrt[3]{x^2(2x - x^2)}\)
\(\frac {dy}{dx} = \frac {10x^{5/3}}{3} - \frac {8x^{2/3}}{3}\)
\(\frac {dy}{dx} = \frac {10x^{5/3}}{3} - \frac {8x^{5/3}}{3}\)
\(\frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{2/3}}{3}\)
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The differentiation of the given function \(y = \sqrt[3]{x^2(2x - x^2)}\) involves the product and chain rules. Let's break down the solution:
\[y = \sqrt[3]{x^2(2x - x^2)}\]
We can rewrite this as:
\[y = x^{2/3}(2x - x^2)^{1/3}\]
Now, applying the product rule \((uv)' = u'v + uv'\) and chain rule \((g(h(x)))' = g'(h(x)) \cdot h'(x)\), we get:
\[y' = \frac{2}{3}x^{-1/3}(2x - x^2)^{1/3} + x^{2/3} \cdot \frac{1}{3}(2x - x^2)^{-2/3}(2 - 2x)\]
Simplify the expression:
\[y' = \frac{2}{3x^{1/3}}(2x - x^2)^{1/3} - \frac{2x^{2/3}(2 - 2x)}{3(2x - x^2)^{2/3}}\]
Now, let's factor out common terms:
\[y' = \frac{2x^{2/3}}{3(2x - x^2)^{2/3}} \left(3 - 3x - x^{1/3}(2x - x^2)^{1/3}\right)\]
The simplified expression for the derivative is:
\[y' = \frac{2x^{2/3}}{3(2x - x^2)^{2/3}} \left(3 - 3x - x^{1/3}(2x - x^2)^{1/3}\right)\]
Comparing with the given options, it seems there might be a mistake in the assumed correct option. The correct differentiation seems to be:
\[y' = \frac{2x^{2/3}}{3(2x - x^2)^{2/3}} \left(3 - x(2x - x^2)^{1/3}\right)\]
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