find the value of $(f\\circ g)$ at the given value.\n$f(u)=u^{7}-3$, $u = g(x)=\\sqrt{x}$, $x =…

Explanation:

Step1: Apply chain rule

The chain rule states that ((f\circ g)'(x)=f'(g(x))\cdot g'(x)). First, find (f'(u)) and (g'(x)). For (f(u) = u^{7}-3), using the power rule ((x^{n})'=nx^{n - 1}), we have (f'(u)=7u^{6}). For (g(x)=\sqrt{x}=x^{\frac{1}{2}}), using the power rule, (g'(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}).

Step2: Substitute (u = g(x)) into (f'(u))

(f'(g(x))=7(g(x))^{6}=7(\sqrt{x})^{6}=7x^{3}).

Step3: Calculate ((f\circ g)'(x))

((f\circ g)'(x)=f'(g(x))\cdot g'(x)=7x^{3}\cdot\frac{1}{2\sqrt{x}}). Simplify (7x^{3}\cdot\frac{1}{2\sqrt{x}}=\frac{7}{2}x^{3-\frac{1}{2}}=\frac{7}{2}x^{\frac{5}{2}}).

Step4: Evaluate at (x = 1)

Substitute (x = 1) into ((f\circ g)'(x)): ((f\circ g)'(1)=\frac{7}{2}(1)^{\frac{5}{2}}).

Answer:

(\frac{7}{2})