Question

If U and V depend on x as U = x 2 + 4x and V = x 3 – 12x, then the value of dV/dU is
Options:

Answer 1

$\implies\bf U = x^2 + 4x$

$\to\rm dU = 2x + 4=2(x+2)$

$\implies\bf V = x^3 -12x$

$\to\rm dV = 3x^2- 12 = 3(x^2-4)$

$\therefore a^2 - b^2 = (a+b)(a-b)$

$\to\rm dV = 3(x^2-4) =3(x^2-2^2)$

$\to\rm dV = 3(x+2)(x-2)$

$\implies\rm \dfrac{dV}{dU} = \dfrac{3(x+2)(x-2)}{2(x+2)}$

$\to\rm \dfrac{dV}{dU} = \dfrac{3(x-2)}{2}$

$\star\bf \dfrac{dV}{dU} = \dfrac{3}{2} \bigg(x-2\bigg)$

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Answer 2

Explanation:

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$\large\bf{\underline{\red{VERIFIED✔}}}$

$\implies\bf U = x^2 + 4x$

$\to\rm dU = 2x + 4=2(x+2)$

$\implies\bf V = x^3 -12x$

$\to\rm dV = 3x^2- 12 = 3(x^2-4)$

$\therefore a^2 - b^2 = (a+b)(a-b)$

$\to\rm dV = 3(x^2-4) =3(x^2-2^2)$

$\to\rm dV = 3(x+2)(x-2)$

$\implies\rm \dfrac{dV}{dU} = \dfrac{3(x+2)(x-2)}{2(x+2)}$

$\to\rm \dfrac{dV}{dU} = \dfrac{3(x-2)}{2}$

$\star\bf \dfrac{dV}{dU} = \dfrac{3}{2} \bigg(x-2\bigg)$

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