y=cos−1⁡3cos⁡x+4sin⁡x5 then dydx equals

y=cos−135cosx+45sinx cosθ=35, sinθ=45y=cos−1⁡(cos⁡(x−θ))=x−θdydx=1−0⇒dydx=1

y=cos−1⁡3cos⁡x+4sin⁡x5 then dydx equals

y=cos−135cosx+45sinx cosθ=35, sinθ=45y=cos−1⁡(cos⁡(x−θ))=x−θdydx=1−0⇒dydx=1

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$y = \cos^{- 1} [\frac{3 \cos x + 4 \sin x}{5}] then \frac{d y}{d x} equals$

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$\frac{1}{\sqrt{1 - x^{3}}}$

$\frac{1}{\sqrt{1 + x^{3}}}$

answer is B.

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$y = \cos^{- 1} [\frac{3}{5} \cos x + \frac{4}{5} \sin x]$ cos $θ$ = $\frac{3}{5}, \sin θ = \frac{4}{5}$

$\begin{matrix} y = \cos^{- 1} (\cos (x - θ)) = x - θ \\ \frac{d y}{d x} = 1 - 0 \Rightarrow \frac{d y}{d x} = 1 \end{matrix}$

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y=cos−1⁡3cos⁡x+4sin⁡x5 then dydx equals

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$y = \cos^{- 1} [\frac{3 \cos x + 4 \sin x}{5}] then \frac{d y}{d x} equals$