## University Calculus: Early Transcendentals (3rd Edition)

$-\dfrac{2e^{\cos (\ln x^2)}}{x}$
Here, $y'=-e^{\cos (\ln x^2)}(\dfrac{d (\ln x^2)}{dx}$ We need to apply the chain rule: Then $y'=-e^{\cos (\ln x^2)} (\dfrac{1}{x^2}) (2x)$ or, $y'=-\dfrac{2e^{\cos (\ln x^2)}}{x}$