If f'(x)=\cos(\ln x) and y=f\left(\frac{2x-3}{x}\right), then what is \frac{dy}{dx} equal to?

Explanation

By the chain rule, \frac{dy}{dx} = f'\left(\frac{2x-3}{x}\right) \cdot \frac{d}{dx}\left(\frac{2x-3}{x}\right). Given f'(x) = \cos(\ln x), we have f'\left(\frac{2x-3}{x}\right) = \cos\left(\ln\left(\frac{2x-3}{x}\right)\right). Now, differentiate the inner function: \frac{d}{dx}\left(2 - 3x^{-1}\right) = 0 - 3(-1x^{-2}) = \frac{3}{x^2}. Multiplying these together gives \frac{3}{x^2}\cos\left(\ln\left(\frac{2x-3}{x}\right)\right).

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