How do you use the chain rule to differentiate \[{\log _{13}}(8{x^3} + 8)\] ?

Hint: First we need to change the base of the logarithm, then we need to apply the chain rule. We know the chain rule that is \[\dfrac{d}{{dx}}(f(g(x)) = f'(g(x)).g'(x)\] . We also know the formula of change of base (natural log) that is \[{\log _a}b = \dfrac{{\ln b}}{{\ln a}}\] . Applying this we can find the required result.

Complete step-by-step answer:
Given,
 \[{\log _{13}}(8{x^3} + 8)\] .
Now applying the formula,
 \[{\log _a}b = \dfrac{{\ln b}}{{\ln a}}\] , where \[a = 13,b = (8{x^3} + 8)\] .
 \[{\log _{13}}(8{x^3} + 8) = \dfrac{{\ln \left( {8{x^3} + 8} \right)}}{{\ln (13)}}\] .
Now we have chain rule that is
 \[\dfrac{d}{{dx}}(f(g(x)) = f'(g(x)).g'(x){\text{ }} - - - (1)\]
Here \[f(x) = \ln x\] , differentiate with respect to ‘x’.
 \[f'(x) = \dfrac{1}{x}\]
Here \[g(x) = 8{x^3} + 8\] , differentiate with respect to ‘x’.
 \[g'(x) = 24{x^2}\]
Then \[f'(g(x)) = \dfrac{1}{{(8{x^3} + 8)}}\] . Substituting these in equation (1) we have,
We have \[\dfrac{1}{{\ln (13)}}\] is constant we take it outside
 \[\dfrac{d}{{dx}}(f(g(x)) = \dfrac{1}{{\ln (13)}}\dfrac{1}{{(8{x^3} + 8)}}.24{x^2}\]
 \[\dfrac{d}{{dx}}(f(g(x)) = \dfrac{1}{{\ln (13)}}\dfrac{1}{{({x^3} + 1)8}}.24{x^2}\]
 \[\dfrac{d}{{dx}}(f(g(x)) = \dfrac{1}{{\ln (13)}}\dfrac{1}{{({x^3} + 1)}}.3{x^2}\]
Thus we have,
 \[ \Rightarrow \dfrac{d}{{dx}}(f(g(x)) = \dfrac{{3{x^2}}}{{\ln (13)({x^3} + 1)}}\] . This is the required answer.
So, the correct answer is “ \[ \Rightarrow \dfrac{d}{{dx}}(f(g(x)) = \dfrac{{3{x^2}}}{{\ln (13)({x^3} + 1)}}\] ”.

Note: We know the differentiation of \[{x^n}\] with respect to ‘x’ is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\] . The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying product rule. We also know that differentiation of constant terms is zero.