Question-140294

Question Number 140294 by solihin last updated on 06/May/21

Answered by Satyendra last updated on 06/May/21

c) \sin^{- 1} y^{2} = \sin h (x y)

\Rightarrow \frac{2 y}{\sqrt{1 - y^{4}}} y^{'} = \cos h (x y) (y + {x y}^{'})

\Rightarrow \frac{2 y}{\sqrt{1 - y^{4}}} y^{'} = y \cos h (x y) + {x y}^{'} \cos h (x y)

\Rightarrow \frac{2 y}{\sqrt{1 - y^{4}}} y^{'} - {x y}^{'} \cosh (x y) = y \cosh (x y)

\Rightarrow y^{'} [\frac{2 y}{\sqrt{1 - y^{4}}} - x \cosh (x y)] = y \cosh (x y)

\Rightarrow y^{'} = \frac{y \cosh (x y)}{\frac{2 y}{\sqrt{1 - y^{4}}} - x \cosh (x y)}

\Rightarrow y^{'} = \frac{y \cosh (x y) \sqrt{1 - y^{4}}}{2 y - x \cosh (x h) \sqrt{1 - y^{4}}}

\Rightarrow \frac{d y}{d x} = \frac{y \cosh (x y) \sqrt{1 - y^{4}}}{2 y - x \cosh (x h) \sqrt{1 - y^{4}}}

Answered by Satyendra last updated on 06/May/21

(a) y = {(\cos^{- 1} 4 x^{2})}^{5}

\Rightarrow \frac{d y}{d x} = \frac{d}{d x} {(\cos^{- 1} 4 x^{2})}^{5}

\Rightarrow \frac{d y}{d x} = 5 {(\cos^{- 1} 4 x^{2})}^{4} \frac{d}{d x} (\cos^{- 1} 4 x^{2})

\Rightarrow \frac{d y}{d x} = 5 {(\cos^{- 1} 4 x^{2})}^{4} \frac{1}{- \sqrt{1 - {(4 x^{2})}^{2}}} \frac{d}{d x} (4 x^{2})

\Rightarrow \frac{d y}{d x} = 5 {(\cos^{- 1} 4 x^{2})}^{4} \frac{1}{- \sqrt{1 - 16 x^{4}}} 8 x

\Rightarrow \frac{d y}{d x} = - \frac{40 x {(\cos^{- 1} 4 x^{2})}^{4}}{\sqrt{1 - 16 x^{4}}}

Answered by liberty last updated on 06/May/21

(a) y = {(\cos^{- 1} {4 x}^{2})}^{5}

\Rightarrow y^{\frac{1}{5}} = \cos^{- 1} ({4 x}^{2})

\Rightarrow {4 x}^{2} = \cos (y^{\frac{1}{5}})

\Rightarrow 8 x = \frac{1}{5} y^{- \frac{4}{5}} (- \sin y^{\frac{1}{5}}) . \frac{dy}{dx}

\Rightarrow \frac{dy}{dx} = - \frac{40 x}{y^{- \frac{4}{5}} (\sin y^{\frac{1}{5}})}

\Rightarrow \frac{dy}{dx} = - 40 x \csc (\cos^{- 1} ({4 x}^{2})) . {(\cos^{- 1} ({4 x}^{2}))}^{4}

Answered by qaz last updated on 06/May/21

≪ c ≫

\frac{2 y}{\sqrt{1 - y^{2}}} y^{'} = \cosh (x y) [y + {x y}^{'}] = y \cosh (x y) + {x y}^{'} \cosh (x y)

\Rightarrow y^{'} = \frac{y \cosh (x y)}{\frac{2 y}{\sqrt{1 - y^{2}}} - x \cosh (x y)}

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