If 2f(sin x) + f(cos x) = x, then frac{d}{dx} | vedclass

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(C) Given: $2f(\sin x) + f(\cos x) = x$  --- $(1)$Replace $x$ with $\frac{\pi}{2} - x$:$2f(\sin(\frac{\pi}{2} - x)) + f(\cos(\frac{\pi}{2} - x)) = \fr

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$\frac{1}{\sqrt{1 - x^2}}$

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(C) Given: $2f(\sin x) + f(\cos x) = x$  --- $(1)$Replace $x$ with $\frac{\pi}{2} - x$:$2f(\sin(\frac{\pi}{2} - x)) + f(\cos(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$$2f(\cos x) + f(\sin x) = \frac{\pi}{2} - x$  --- $(2)$Multiply equation $(2)$ by $2$:$4f(\cos x) + 2f(\sin x) = \pi - 2x$  --- $(3)$Subtract equation $(1)$ from equation $(3)$:$(4f(\cos x) + 2f(\sin x)) - (2f(\sin x) + f(\cos x)) = (\pi - 2x) - x$$3f(\cos x) = \pi - 3x$$f(\cos x) = \frac{\pi}{3} - x$Let $t = \cos x$,then $x = \cos^{-1} t$. Substituting this:$f(t) = \frac{\pi}{3} - \cos^{-1} t$Thus,$f(x) = \frac{\pi}{3} - \cos^{-1} x$Differentiating with respect to $x$:$f'(x) = \frac{d}{dx}(\frac{\pi}{3} - \cos^{-1} x) = 0 - (-\frac{1}{\sqrt{1 - x^2}}) = \frac{1}{\sqrt{1 - x^2}}$

If $2f(\sin x) + f(\cos x) = x,$ then $\frac{d}{dx} f(x)$ is

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