Categories > differentiation > Differentiation by substitution > Let $f(x) = \begin{vmatrix} {{x^3}}&{\sin x}&{\cos x}\\ 6&{ - 1}&0\\ p&{{p^2}}&{{p^3}} \end{vmatrix}$, where $p$ is a constant. Then $\frac{{{d}^{3}}}{d{{x}^{3}}}\left\{ f(x) \right\}$at $x=0$is

differentiation

Question: Let $f(x) = \begin{vmatrix} {{x^3}}&{\sin x}&{\cos x}\\ 6&{ - 1}&0\\ p&{{p^2}}&{{p^3}} \end{vmatrix}$, where $p$ is a constant. Then $\frac{{{d}^{3}}}{d{{x}^{3}}}\left\{ f(x) \right\}$at $x=0$is


1) $p$
2) $p+{{p}^{2}}$
3) $p+{{p}^{3}}$
4) Independent of $p$
Solution: Explanation: No Explanation
Differentiation by substitution Higher order derivatives

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