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  • Help me understand! - Limit , continuity and differentiability - JEE Main-9

Let f(x) = \begin{vmatrix} x^{3} & 1 &3 \\ x^{4}&2 & 4\\ x^{2} & 3 & 2 \end{vmatrix} then f'(1) = ?

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Answers (1)

best_answer

As we have learnt,

 

Differentiation of a determinant -

y=\begin{vmatrix} f(x) & g(x) & h(x) \\ p(x)& q(x) & r(x) \\ u(x)& v(x)& w(x)) \end{vmatrix} then 

\frac{dy}{dx}=\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ p(x)& q(x) & r(x) \\ u(x)& v(x)& w(x)) \end{vmatrix}+\begin{vmatrix} f(x) & g(x) & h(x) \\ p'(x)& q'(x) & r'(x) \\ u(x)& v(x)& w(x)) \end{vmatrix}+\begin{vmatrix} f(x) & g(x) & h(x) \\ p(x)& q(x) & r(x) \\ u'(x)& v'(x)& w'(x)) \end{vmatrix}

 


 

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 Differciating Columnwise,

f'(x) = \begin{vmatrix} 3x^2 &1 &3 \\ 4x^3& 2& 4\\ 2x&3 & 2 \end{vmatrix} + \begin{vmatrix} x^3 & 0 &3 \\ x^4 & 0 &4 \\ x^2 & 0 & 2 \end{vmatrix} + \begin{vmatrix} x^3 & 1 &0 \\ x^4 & 2 &0 \\ x^2 & 3 & 0 \end{vmatrix}

f'(1) = \begin{vmatrix} 3 &1 &3 \\ 4& 2& 4\\ 2&3 & 2 \end{vmatrix} + 0 + 0 = 0 + 0 + 0 = 0 \;\;\;\;(\because C_1 = C_3)

 


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