Differentiation Question 400
Question: $ f(x)= \begin{vmatrix} x^{3} & x^{2} & 3x^{2} \\ 1 & -6 & 4 \\ p & p^{2} & p^{3} \\ \end{vmatrix} $ , here p is a constant, then $ \frac{d^{3}f(x)}{dx^{3}} $ is
[DCE 2000]
Options:
A) Proportional to $ x^{2} $
B) Proportional to x
C) Proportional to $ x^{3} $
D) A constant
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(x)=| \begin{matrix} x^{3} & x^{2} & 3x^{2} \\ 1 & -6 & 4 \\ p & p^{2} & p^{3} \\ \end{matrix} | $
Therefore $ f(x)=x^{3}(-6p^{3}-4p^{2})-x^{2}(p^{3}-4p)+3x^{2}(p^{2}+6p) $
Therefore $ f(x)=-6p^{3}x^{3}-4p^{2}x^{3}-x^{2}p^{3}+4px^{2}+3p^{2}x^{2}+18px^{2} $ \ $ \frac{d}{dx}f(x)=-18p^{3}x^{2}-12p^{2}x^{2}-2xp^{3}+8px+6p^{2}x+36px $
and $ \frac{d^{2}}{dx^{2}}f(x)=-36p^{3}x-24p^{2}x-2p^{3}+8p+6p^{2}+36p $
and $ \frac{d^{3}f(x)}{dx^{3}}=-36p^{3}-24p^{2} $ = a constant.
BETA
