If 3 f(cos x) + 2 f(sin x) = 5 x,then f^{prim | vedclass

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(C) Given: $3 f(\cos x) + 2 f(\sin x) = 5 x$  ...$(i)$Replace $x$ with $(\frac{\pi}{2} - x)$:$3 f(\sin x) + 2 f(\cos x) = 5(\frac{\pi}{2} - x)$  ...$(

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$\frac{-5}{\sin x} - \frac{5}{\cos x}$

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(C) Given: $3 f(\cos x) + 2 f(\sin x) = 5 x$  ...$(i)$Replace $x$ with $(\frac{\pi}{2} - x)$:$3 f(\sin x) + 2 f(\cos x) = 5(\frac{\pi}{2} - x)$  ...$(ii)$Multiply $(i)$ by $3$ and $(ii)$ by $2$:$9 f(\cos x) + 6 f(\sin x) = 15 x$$4 f(\cos x) + 6 f(\sin x) = 10(\frac{\pi}{2} - x) = 5\pi - 10 x$Subtracting the equations:$5 f(\cos x) = 25 x - 5\pi \Rightarrow f(\cos x) = 5 x - \pi$Differentiating with respect to $x$:$f^{\prime}(\cos x) \cdot (-\sin x) = 5 \Rightarrow f^{\prime}(\cos x) = \frac{-5}{\sin x}$Similarly,from $(i)$ and $(ii)$,we find $f(\sin x) = \frac{5\pi}{2} - 5x$:$f^{\prime}(\sin x) \cdot (\cos x) = -5 \Rightarrow f^{\prime}(\sin x) = \frac{-5}{\cos x}$Therefore,$f^{\prime}(\cos x) + f^{\prime}(\sin x) = \frac{-5}{\sin x} - \frac{5}{\cos x}$.

If $3 f(\cos x) + 2 f(\sin x) = 5 x$,then $f^{\prime}(\cos x) + f^{\prime}(\sin x) =$

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