If y=5 sin x+6 cos x,then y^2+(y_1)^2= . . . | vedclass

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(A) Given $y = 5 \sin x + 6 \cos x$. First,find the derivative $y_1 = \frac{dy}{dx} = 5 \cos x - 6 \sin x$. Now,calculate $y^2 + (y_1)^2$: $y^2 = (5 \

Vedclass · Accepted Answer

$61$

Vedclass · Answer

(A) Given $y = 5 \sin x + 6 \cos x$. First,find the derivative $y_1 = \frac{dy}{dx} = 5 \cos x - 6 \sin x$. Now,calculate $y^2 + (y_1)^2$: $y^2 = (5 \sin x + 6 \cos x)^2 = 25 \sin^2 x + 36 \cos^2 x + 60 \sin x \cos x$. $(y_1)^2 = (5 \cos x - 6 \sin x)^2 = 25 \cos^2 x + 36 \sin^2 x - 60 \sin x \cos x$. Adding these two expressions: $y^2 + (y_1)^2 = (25 \sin^2 x + 36 \cos^2 x + 60 \sin x \cos x) + (25 \cos^2 x + 36 \sin^2 x - 60 \sin x \cos x)$. $y^2 + (y_1)^2 = 25(\sin^2 x + \cos^2 x) + 36(\cos^2 x + \sin^2 x)$. Since $\sin^2 x + \cos^2 x = 1$,we get: $y^2 + (y_1)^2 = 25(1) + 36(1) = 61$.

If $y=5 \sin x+6 \cos x$,then $y^2+(y_1)^2=$ . . . . . .

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