Solved on Nov 06, 2023

Find the derivative of q(x)=log8(6x3+3x+1)q(x)=\log_{8}(6x^{3}+3x+1).

STEP 1

Assumptions1. We are given the function q(x)=log8(6x3+3x+1)q(x)=\log{8}\left(6 x^{3}+3 x+1\right). We need to find the derivative of this function, denoted as q(x)q^{\prime}(x)3. We will use the chain rule and the rule for the derivative of a logarithm to solve this problem

STEP 2

The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In this case, our outer function is the logarithm and our inner function is 6x+x+16x^ +x +1.

STEP 3

The derivative of a logarithm base bb is given by the formuladdxlogb(u)=1uln(b)u\frac{d}{dx} \log_b(u) = \frac{1}{u \ln(b)} \cdot u'where uu is a function of xx and uu' is the derivative of uu.

STEP 4

We first find the derivative of the inner function u=6x3+3x+1u =6x^3 +3x +1.
u=ddx(6x3+3x+1)u' = \frac{d}{dx} (6x^3 +3x +1)

STEP 5

Calculate the derivative of the inner function.
u=18x2+3u' =18x^2 +3

STEP 6

Now we can find the derivative of the original function q(x)q(x) using the chain rule and the formula for the derivative of a logarithm.
q(x)=16x3+3x+11ln(8)(18x2+3)q^{\prime}(x) = \frac{1}{6x^3 +3x +1} \cdot \frac{1}{\ln(8)} \cdot (18x^2 +3)

STEP 7

implify the expression.
q(x)=18x2+3(6x3+3x+1)ln()q^{\prime}(x) = \frac{18x^2 +3}{(6x^3 +3x +1) \ln()}So, the derivative of the function q(x)=log(6x3+3x+1)q(x)=\log{}\left(6 x^{3}+3 x+1\right) is q(x)=18x2+3(6x3+3x+1)ln()q^{\prime}(x) = \frac{18x^2 +3}{(6x^3 +3x +1) \ln()}.

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