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astraxan [27]
2 years ago
13

Derive this with respect to x

Formula1" title=" \frac{3}{(1 + secx) } " alt=" \frac{3}{(1 + secx) } " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Elenna [48]2 years ago
8 0

~~~~\dfrac{d}{dx} \left(\dfrac{3}{1 + \sec x} \right)\\\\\\=3 \dfrac{d}{dx} \left( \dfrac 1{ 1+ \sec x} \right)\\\\\\=3 \dfrac{d}{dx} (1+ \sec x)^{-1}\\\\\\=3 (-1) (1 + \sec x )^{-1 -1} \dfrac{d}{dx}( 1 + \sec x)\\\\\\=-3(1 + \sec x)^{-2}  ( 0 + \sec x  \tan x)\\\\\\=-\dfrac{3\sec x \tan x}{(1 + \sec x)^2}

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Irina18 [472]

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a. \quad\dfrac{\left|\begin{array}{cc}148&14\\246&5\end{array}\right|}{\left|\begin{array}{cc}-2&14\\3&5\end{array}\right|}=52

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To find the value of x, Cramer's rule says you replace the x-coefficients with the equation constants to form the matrix whose determinant is the numerator of the fraction. The denominator is the determinant of the matrix of coefficients. The equation constants are 148 and 246, so you expect to find those in the first column of the numerator (answer choices A and C).

\dfrac{\left|\begin{array}{cc}148&14\\246&5\end{array}\right|}{\left|\begin{array}{cc}-2&14\\3&5\end{array}\right|}=\dfrac{-2704}{-52}=52

The calculation is carried out correctly only in answer choice A.

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