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Differentiation of a Determinant

Let f x =co...

Question

Let $f (x) = ∣ ∣ ∣ ∣ \begin{matrix} cos x & sin x & cos x cos 2 x & sin 2 x & 2 cos 2 x cos 3 x & sin 3 x & 3 cos 3 x \end{matrix} ∣ ∣ ∣ ∣$
Then

A

f^{'} (0) = 0

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B

f^{'} (π / 2) = 4

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C

f^{'} (π) = 0

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D

f^{'} (π) = 1

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Solution

The correct options are
A $f^{'} (0) = 0$
B $f^{'} (π / 2) = 4$
C $f^{'} (π) = 0$
$f (x) = ∣ ∣ ∣ ∣ \begin{matrix} cos x & sin x & cos x cos 2 x & sin 2 x & 2 cos 2 x cos 3 x & sin 3 x & 3 cos 3 x \end{matrix} ∣ ∣ ∣ ∣ \Rightarrow f^{'} (x) = ∣ ∣ ∣ ∣ \begin{matrix} - sin x & sin x & cos x - 2 sin 2 x & sin 2 x & 2 cos 2 x - 3 sin 3 x & sin 3 x & 3 cos 3 x \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} cos x & cos x & cos x cos 2 x & 2 cos 2 x & 2 cos 2 x cos 3 x & 3 cos 3 x & 3 cos 3 x \end{matrix} ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ \begin{matrix} cos x & sin x & - sin x cos 2 x & sin 2 x & - 4 sin 2 x cos 3 x & sin 3 x & - 9 sin 3 x \end{matrix} ∣ ∣ ∣ ∣ ∴ f^{'} (0) = 0 = f^{'} (π), f^{'} (\frac{π}{2}) = 4$

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Similar questions

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} cos x & sin x & cos x cos 2 x & sin 2 x & 2 cos 2 x cos 3 x & sin 3 x & 3 cos 3 x \end{matrix} ∣ ∣ ∣ ∣

f^{'} (\frac{π}{2}) =

.

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} cos x & sin x & cos x cos 2 x & sin 2 x & 2 cos 2 x cos 3 x & sin 3 x & 3 cos 3 x \end{matrix} ∣ ∣ ∣ ∣

f^{'} (\frac{π}{2}) =

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} c o s x & s i n x & c o s x c o s 2 x & s i n 2 x & 2 c o s 2 x c o s 3 x & s i n 3 x & 3 c o s 3 x \end{matrix} ∣ ∣ ∣ ∣

, then the value of

f^{'} (π / 2) =

.....................

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} c o s x & s i n x & c o s x c o x 2 x & s i n 2 x & 2 c o s 2 x c o s 3 x & s i n 3 x & 3 c o s 3 x \end{matrix} ∣ ∣ ∣ ∣

f^{'} (\frac{π}{2})

f (x) = | cos 2 x |

f^{^{'}} (\frac{π}{4} + 0)

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Differentiation of a Determinant

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