Question

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

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.

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Answer:

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