Differentiate the function with respect to x: | vedclass

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(B) Let $y = \cos (a \cos x + b \sin x)$.By using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} [\cos (a \cos x +

Vedclass · Accepted Answer

$(a \sin x - b \cos x) \sin (a \cos x + b \sin x)$

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(B) Let $y = \cos (a \cos x + b \sin x)$.By using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} [\cos (a \cos x + b \sin x)]$Using the derivative of $\cos(u)$ which is $-\sin(u) \cdot \frac{du}{dx}$:$\frac{dy}{dx} = -\sin (a \cos x + b \sin x) \cdot \frac{d}{dx} (a \cos x + b \sin x)$Now,differentiate the inner function:$\frac{d}{dx} (a \cos x + b \sin x) = a(-\sin x) + b(\cos x) = b \cos x - a \sin x$Substituting this back into the expression:$\frac{dy}{dx} = -\sin (a \cos x + b \sin x) \cdot (b \cos x - a \sin x)$Rearranging the terms:$\frac{dy}{dx} = (a \sin x - b \cos x) \sin (a \cos x + b \sin x)$

Differentiate the function with respect to $x$: $\cos (a \cos x + b \sin x)$,where $a$ and $b$ are constants.

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