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