find the derivative.\n$\frac{d}{dx}3x(x^{6}+1)^{8}$\n$\frac{d}{dx}3x(x^{6}+1)^{8}=square$

Explanation:

Step1: Apply product - rule

The product - rule states that if (y = uv), where (u) and (v) are functions of (x), then (y^\prime=u^\prime v + uv^\prime). Let (u = 3x) and (v=(x^{6}+1)^{8}). First, find (u^\prime) and (v^\prime). The derivative of (u = 3x) with respect to (x) is (u^\prime=\frac{d}{dx}(3x)=3).

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

The chain - rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Let (g(x)=x^{6}+1) and (f(u)=u^{8}). Then (g^\prime(x)=\frac{d}{dx}(x^{6}+1)=6x^{5}) and (f^\prime(u) = 8u^{7}). Substituting (u = x^{6}+1) back into (f^\prime(u)), we get (f^\prime(g(x))=8(x^{6}+1)^{7}). So (v^\prime=\frac{d}{dx}(x^{6}+1)^{8}=8(x^{6}+1)^{7}\cdot6x^{5}=48x^{5}(x^{6}+1)^{7}).

Step3: Calculate the derivative using product - rule

Using the product - rule (y^\prime=u^\prime v+uv^\prime), we substitute (u = 3x), (u^\prime = 3), (v=(x^{6}+1)^{8}), and (v^\prime=48x^{5}(x^{6}+1)^{7}) into it. [ \begin{align*} \frac{d}{dx}[3x(x^{6}+1)^{8}]&=3(x^{6}+1)^{8}+3x\cdot48x^{5}(x^{6}+1)^{7}\ &=3(x^{6}+1)^{8}+144x^{6}(x^{6}+1)^{7}\ &=3(x^{6}+1)^{7}[(x^{6}+1)+48x^{6}]\ &=3(x^{6}+1)^{7}(49x^{6}+1) \end{align*} ]

Answer:

(3(x^{6}+1)^{7}(49x^{6}+1))