Let f x - begin vmatrix cos x -sin x -cos xcox 2 x -sin 2 x - 2 cos 2 xcos 3 x -sin 3 x - 3 cos 3 xlendvmatrix - then f-2 is equal toA- 4B- 8C- 2D- 6

The correct option is A 4f'(x)= sin x amp; sin x amp; cosx 2 sin 2x amp; sin 2x amp; 2 cos 2x 3 sin 3x amp; sin 3x amp; 3 cos 3x + c ...

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Differentiation of a Determinant

Let f x =⟨ be...

Question

Let $f (x) = ∣ ∣ ∣ ∣ \begin{matrix} c o s x & s i n x & c o s x c o x 2 x & s i n 2 x & 2 c o s 2 x c o s 3 x & s i n 3 x & 3 c o s 3 x \end{matrix} ∣ ∣ ∣ ∣$ , then $f^{'} (\frac{π}{2})$ is equal to

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Solution

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Similar questions

If f (x) = | cos x - sin x |

f^{'} (\frac{π}{6})

f (x) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 c o s^{2} x & S i n 2 x & - S i n x S i n 2 x & 2 s i n^{2} x & C o s x S i n x & - C o s x & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

\int_{0}^{π / 2} [f (x) + f^{1} (x)] d x =

\cos x + \cos 2 x + \cos 3 x = 0

\cos x + \cos 3 x - \cos 2 x = 0

\sin x + \sin 5 x = \sin 3 x

\cos x \cos 2 x \cos 3 x = \frac{1}{4}

\cos x + \sin x = \cos 2 x + \sin 2 x

\sin x + \sin 2 x + \sin 3 = 0

\sin x + \sin 2 x + \sin 3 x + \sin 4 x = 0

\sin 3 x - \sin x = 4 \cos^{2} x - 2

\sin 2 x - \sin 4 x + \sin 6 x = 0

f (x) = {∣ ∣ ∣ ∣ \begin{matrix} 0 & cos x & - sin x sin x & 0 & cos x cos x & sin x & 0 \end{matrix} ∣ ∣ ∣ ∣}^{2}

sin 2 x = 1

f (x) = 2 / 3

f (x) = 0

sin x = cos x

Δ (x)

∣ ∣ ∣ ∣ ∣ \begin{matrix} c o s^{2} x & c o s x s i n x & - s i n x c o s x s i n x & s i n^{2} x & c o s x s i n x & - c o s x & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

\int_{0}^{π / 2} [Δ (x) + Δ^{1} (x)] d x

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