Solved on Nov 06, 2023

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

STEP 1

Assumptions1. We are given the function $q(x)=\log{8}\left(6 x^{3}+3 x+1\right)$ . We need to find the derivative of this function, denoted as $q^{\prime}(x)$ 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 $6x^ +x +1$ .

STEP 3

The derivative of a logarithm base $b$ is given by the formula $\frac{d}{dx} \log_b(u) = \frac{1}{u \ln(b)} \cdot u'$ where $u$ is a function of $x$ and $u'$ is the derivative of $u$ .

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

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

STEP 1

Assumptions1. We are given the function $q(x)=\log{8}\left(6 x^{3}+3 x+1\right)$ . We need to find the derivative of this function, denoted as $q^{\prime}(x)$ 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 $6x^ +x +1$ .

STEP 3

The derivative of a logarithm base $b$ is given by the formula $\frac{d}{dx} \log_b(u) = \frac{1}{u \ln(b)} \cdot u'$ where $u$ is a function of $x$ and $u'$ is the derivative of $u$ .

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

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

STEP 1

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 +16x+x+1.

STEP 3

The derivative of a logarithm base bbb is given by the formuladdxlog⁡b(u)=1uln⁡(b)⋅u′\frac{d}{dx} \log_b(u) = \frac{1}{u \ln(b)} \cdot u'dxd​logb​(u)=uln(b)1​⋅u′where uuu is a function of xxx and u′u'u′ is the derivative of uuu.

STEP 4

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

STEP 5

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

STEP 6

STEP 7

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

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 $6x^ +x +1$ .

The derivative of a logarithm base $b$ is given by the formula $\frac{d}{dx} \log_b(u) = \frac{1}{u \ln(b)} \cdot u'$ where $u$ is a function of $x$ and $u'$ is the derivative of $u$ .

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

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