differentiate the following function.\n\n$y = 6e^{x}(3 + 5e^{x})^{2}$\n\n$\frac{dy}{dx} = square$

Explanation:

Step1: Apply the product rule

The product rule states that if (y = uv), then (y^\prime=u^\prime v + uv^\prime). Let (u = 6e^{x}) and (v=(3 + 5e^{x})^{2}). First, find (u^\prime): (u^\prime=\frac{d}{dx}(6e^{x})=6e^{x})

Step2: Apply the chain rule to find (v^\prime)

Let (t = 3+5e^{x}), so (v = t^{2}). By the chain rule (\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}). (\frac{dv}{dt} = 2t) and (\frac{dt}{dx}=5e^{x}). Then (\frac{dv}{dx}=2(3 + 5e^{x})\cdot5e^{x}=10e^{x}(3 + 5e^{x}))

Step3: Substitute (u), (u^\prime), (v), (v^\prime) into the product rule

(y^\prime=u^\prime v+uv^\prime) (y^\prime=6e^{x}(3 + 5e^{x})^{2}+6e^{x}\cdot10e^{x}(3 + 5e^{x})) Factor out (6e^{x}(3 + 5e^{x})): (y^\prime=6e^{x}(3 + 5e^{x})[(3 + 5e^{x})+10e^{x}]) Simplify the expression inside the brackets: ((3 + 5e^{x})+10e^{x}=3 + 15e^{x}=3(1 + 5e^{x})) So (y^\prime=18e^{x}(3 + 5e^{x})(1 + 5e^{x}))

Answer:

(18e^{x}(3 + 5e^{x})(1 + 5e^{x}))