Categories > differentiation > Differentiation by substitution > $f(x) = \begin{vmatrix} {{x^3}}&{{x^2}}&{3{x^2}}\\ 1&{ - 6}&4\\ p&{{p^2}}&{{p^3}} \end{vmatrix}$, here $p$ is a constant, then $\frac{{{d}^{3}}f(x)}{d{{x}^{3}}}$ is

differentiation

Question: $f(x) = \begin{vmatrix} {{x^3}}&{{x^2}}&{3{x^2}}\\ 1&{ - 6}&4\\ p&{{p^2}}&{{p^3}} \end{vmatrix}$, here $p$ is a constant, then $\frac{{{d}^{3}}f(x)}{d{{x}^{3}}}$ is


1) Proportional to ${{x}^{2}}$
2) Proportional to $x$
3) Proportional to ${{x}^{3}}$
4) A constant
Solution: Explanation: No Explanation
Differentiation by substitution Higher order derivatives

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