If - 2 sin x -sin n x - sin x -sin n x - 2 sin x -sin n x - sin x -sin n x - - -x- 2 3- x- 2 where -x- denotes the integral part of x and x- 0- and n- 1 Show that fx is continuous and differentiable at x-2-

We have #160; f( x ) = [ 2( sin x sin^nx) + | sin x sin^nx|/2( sin x sin^nx) | sin x sin^nx|];x π/2 f(x) = 3 ; x = π/6 If x ∈( 0, π ...

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Derivative of Standard Functions

If [ 2 sin ...

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If $⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} [\frac{2 (sin x - {sin}^{n} x) + | sin x - {sin}^{n} x |}{2 (sin x - {sin}^{n} x) - | sin x - {sin}^{n} x |}], x \neq \frac{π}{2} 3, x = \frac{π}{2} \end{matrix}$ where $[x]$ denotes the integral part of $x$ and $x \in (0, π)$ and $n > 1$
Show that $f (x)$ is continuous and differentiable at $x = \frac{π}{2}$ .

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f : R \to R

f (x) = \frac{s i n [x] π + t a n [x] π}{1 + [x^{2}]}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} [\frac{2 (sin x - {sin}^{3} x) + ∣ ∣ sin x - {sin}^{3} x ∣ ∣}{2 (sin x - {sin}^{3} x) - ∣ ∣ sin x - {sin}^{3} x ∣ ∣}], x \neq \frac{π}{2} f o r x \in (0, π) 3 x = \frac{π}{2} \end{matrix}

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\int s i n^{n} (x) d x = \frac{- s i n^{n - 1} (x) . c o s (x)}{n} + \frac{n - 1}{n} \int s i n^{n - 2} (x) d x

\int s i n^{3} (x) d x .

f (x) = {\begin{matrix} m x + 1, & x \leq \frac{π}{2} s i n x + n, & x > \frac{π}{2} \end{matrix}

x = \frac{π}{2}

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} m x + 1, & x \leq \frac{π}{2} sin x + n, & x > \frac{π}{2} \end{matrix}

x = \frac{π}{2}

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