If f(x) = left| begin{array}{ccc} cos x & x & | vedclass

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(A) Given $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \ 2\sin x & x^2 & 2x \ 	an x & x & 1 \end{array} ight|$.Expanding the determinant alon

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Exists and is equal to $-2$

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(A) Given $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \ 2\sin x & x^2 & 2x \ 	an x & x & 1 \end{array} ight|$.Expanding the determinant along the first row:$f(x) = \cos x(x^2 - 2x^2) - x(2\sin x - 2x	an x) + 1(2x\sin x - x^2	an x)$$f(x) = -x^2\cos x - 2x\sin x + 2x^2	an x + 2x\sin x - x^2	an x$$f(x) = x^2	an x - x^2\cos x = x^2(	an x - \cos x)$.Now,differentiate $f(x)$ with respect to $x$:$f'(x) = 2x(	an x - \cos x) + x^2(\sec^2 x + \sin x)$.We need to find $\lim_{x 	o 0} \frac{f'(x)}{x}$:$\lim_{x 	o 0} \frac{2x(	an x - \cos x) + x^2(\sec^2 x + \sin x)}{x}$$= \lim_{x 	o 0} [2(	an x - \cos x) + x(\sec^2 x + \sin x)]$$= 2(0 - 1) + 0(1 + 0) = -2$.Thus,the limit exists and is equal to $-2$.

If $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2\sin x & x^2 & 2x \\ \tan x & x & 1 \end{array} \right|$,then find $\lim_{x \to 0} \frac{f'(x)}{x}$.

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