Question

$$ \frac { d } { d x } left - 4 e ^ { x } - sin ( x ) - 9 ight = $$

Explanation:

Step1: Apply the sum/difference rule

The sum/difference rule of differentiation states that \((u - v - w)'=u'-v'-w'\). Let \(u = - 4e^{x}\), \(v=\sin(x)\), \(w = 9\). Then \(\frac{d}{dx}[-4e^{x}-\sin(x)-9]=\frac{d}{dx}(-4e^{x})-\frac{d}{dx}(\sin(x))-\frac{d}{dx}(9)\)

Step2: Differentiate each term

• For \(y=-4e^{x}\), using the rule \(\frac{d}{dx}(ae^{x})=ae^{x}\) (where \(a = - 4\)), we have \(\frac{d}{dx}(-4e^{x})=-4e^{x}\)
• For \(y = \sin(x)\), using the rule \(\frac{d}{dx}(\sin(x))=\cos(x)\)
• For \(y = 9\) (a constant), using the rule \(\frac{d}{dx}(c)=0\) (where \(c = 9\))

Step3: Combine the results

Substitute the derivatives of each term back into the expression from Step1: \(-4e^{x}-\cos(x)-0=-4e^{x}-\cos(x)\)

Answer:

\(-4e^{x}-\cos(x)\)