Question

if ( f(x)=5sin x + 4cos x ), then
( f(x)=)
( f(1)=)
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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:

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