7. $g(t)=e^{\tan (t^{2}+1)}$, determine $g(t)$

Explanation:

Step1: Apply the chain rule

Let (u = \tan(t^{2}+1)), then (g(t)=e^{u}). The derivative of (y = e^{u}) with respect to (u) is (y'_{u}=e^{u}).

Step2: Differentiate (u) with respect to (t)

Let (v=t^{2}+1), then (u = \tan(v)). The derivative of (u) with respect to (v) is (u'{v}=\sec^{2}(v)), and the derivative of (v) with respect to (t) is (v'{t} = 2t). By the chain rule (u'{t}=u'{v}\cdot v'_{t}=2t\sec^{2}(t^{2}+1)).

Step3: Apply the chain rule for (g(t))

By the chain rule (g'(t)=g'{u}\cdot u'{t}). Substituting (g'{u}=e^{u}) and (u'{t}=2t\sec^{2}(t^{2}+1)) (where (u = \tan(t^{2}+1))), we get (g'(t)=e^{\tan(t^{2}+1)}\cdot2t\sec^{2}(t^{2}+1)).

Answer:

(g'(t)=2t\sec^{2}(t^{2}+1)e^{\tan(t^{2}+1)})