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Derivative from First Principle

If fx=x-a4 ...

Question

If $f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - a)^{4} & (x - a)^{3} & 1 (x - b)^{4} & (x - b)^{3} & 1 (x - c)^{4} & (x - c)^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ then $f^{'} (x) = λ ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - a)^{4} & (x - a)^{2} & 1 (x - b)^{4} & (x - b)^{2} & 1 (x - c) & (x - c)^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ Find the value of $λ$

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Solution

$f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - a)^{4} & (x - a)^{3} & 1 (x - b)^{4} & (x - b)^{3} & 1 (x - c)^{4} & (x - c)^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$∴ f^{'} (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 4 (x - a)^{3} & (x - a)^{3} & 1 4 (x - b)^{3} & (x - b)^{3} & 1 4 (x - c)^{3} & (x - c)^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - a)^{4} & 3 (x - a)^{2} & 1 (x - b)^{4} & 3 (x - b)^{2} & 1 (x - c)^{4} & 3 (x - c)^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - a)^{4} & (x - a)^{3} & 0 (x - b)^{4} & (x - b)^{3} & 0 (x - c)^{4} & (x - c)^{3} & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow f^{'} (x) = 4 ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - a)^{3} & (x - a)^{3} & 1 (x - b)^{3} & (x - b)^{3} & 1 (x - c)^{3} & (x - c)^{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + 3 ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - a)^{4} & (x - a)^{2} & 1 (x - b)^{4} & (x - b)^{2} & 1 (x - c)^{4} & (x - c)^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + 0$

$= 0 + 3 ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - a)^{4} & (x - a)^{2} & 1 (x - b)^{4} & (x - b)^{2} & 1 (x - c)^{4} & (x - c)^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow f^{'} (x) = 3 ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - a)^{4} & (x - a)^{2} & 1 (x - b)^{4} & (x - b)^{2} & 1 (x - c)^{4} & (x - c)^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$∴ λ = 3$

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