if ( f(x)=5sin x + 4cos x ), then\n( f(x)=)\n( f(1)=)\nquestion help: message instructor\nsubmit question…

Explanation:

Step1: Differentiate term - by - term

Use the derivative rules: ((\sin x)^\prime=\cos x) and ((\cos x)^\prime =-\sin x). For (y = 5\sin x+4\cos x), by the sum rule ((u + v)^\prime=u^\prime + v^\prime) (where (u = 5\sin x) and (v = 4\cos x)), we have (y^\prime=(5\sin x)^\prime+(4\cos x)^\prime). Since ((k\cdot f(x))^\prime=k\cdot f^\prime(x)) (where (k) is a constant), ((5\sin x)^\prime = 5\cos x) and ((4\cos x)^\prime=-4\sin x). So (f^\prime(x)=5\cos x-4\sin x).

Step2: Evaluate (f^\prime(x)) at (x = 1)

Substitute (x = 1) into (f^\prime(x)). We know that (f^\prime(1)=5\cos(1)-4\sin(1)). Using a calculator (where (1) is in radians), (\cos(1)\approx0.5403) and (\sin(1)\approx0.8415). Then (f^\prime(1)=5\times0.5403-4\times0.8415=2.7015 - 3.366=-0.6645).

Answer:

(f^\prime(x)=5\cos x - 4\sin x); (f^\prime(1)\approx - 0.6645)