Question

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

Answer 1

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.