Find the derivative of f(x) = sin x + cos x u | vedclass

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(A) By the first principle,the derivative $f'(x)$ is defined as:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$Substituting $f(x) = \sin x + \cos x$:

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$\cos x - \sin x$

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(A) By the first principle,the derivative $f'(x)$ is defined as:$f'(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$Substituting $f(x) = \sin x + \cos x$:$f'(x) = \lim_{h 	o 0} \frac{\sin(x+h) + \cos(x+h) - (\sin x + \cos x)}{h}$Using the sum formulas $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$:$f'(x) = \lim_{h 	o 0} \frac{\sin x \cos h + \cos x \sin h + \cos x \cos h - \sin x \sin h - \sin x - \cos x}{h}$Rearranging terms:$f'(x) = \lim_{h 	o 0} \left[ \sin x \left( \frac{\cos h - 1}{h} ight) + \cos x \left( \frac{\sin h}{h} ight) + \cos x \left( \frac{\cos h - 1}{h} ight) - \sin x \left( \frac{\sin h}{h} ight) ight]$Using standard limits $\lim_{h 	o 0} \frac{\sin h}{h} = 1$ and $\lim_{h 	o 0} \frac{\cos h - 1}{h} = 0$:$f'(x) = \sin x(0) + \cos x(1) + \cos x(0) - \sin x(1)$$f'(x) = \cos x - \sin x$

Find the derivative of $f(x) = \sin x + \cos x$ using the first principle.

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