Lf Fx- x x2 x3- 1 2x 3x2- 0 2 6x - then F-x is equal to

The correct option is D 6x^2 F(x)=|[ x amp; x^2 amp; x^3; 1 amp; 2x amp; 3x^2; 0 amp; 2 amp; 6x ]| = 2x^3 F'(x) ...

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Determinant

lf Fx=|[ x...

Question

lf $F (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & x^{3} 1 & 2 x & 3 x^{2} 0 & 2 & 6 x \end{matrix} ∣ ∣ ∣ ∣ ∣$ , then $F^{'} (x)$ is equal to

6 x^{3}

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x^{3} + 6 x^{2}

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3 x

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6 x^{2}

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Solution

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Similar questions

f (x) = {cot}^{- 1} (\frac{1 - x^{2}}{2 x}) + {cot}^{- 1} (\frac{1 - 3 x^{2}}{3 x - x^{3}}) - {cot}^{- 1} (\frac{1 - 6 x^{2} + x^{4}}{4 x - 4 x^{3}})

F^{'} (x)

f (x) = x^{3} - 6 x^{2} + 9 x + 3

f (2)

f (x) = {\begin{matrix} x^{3} - 3 x + 2 x < 2 x^{3} - 6 x^{2} + 9 x + 2, x \geq 2 \end{matrix}

f (x) = {\begin{matrix} x^{3} - 3 x + 2, & x < 2 x^{3} - 6 x^{2} + 9 x + 2, & x \geq 2 \end{matrix}

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