Question

Differentiate the following function:
${e^{\log \left( {\log \,\,x} \right)}}.\,\,\log 3x$.

Answer 1

Hint : In this type of problem we first simplify exponential term having logarithmic function in power by using formula ${e^{\log (A)}} = A$ , and then using product rule of differentiation and chain rule of differentiation to get required solution of the given problem.
Formulas Used: Product rule of differentiation $y = u.v,\,\,\,\,\dfrac{{dy}}{{dx}} = u.\dfrac{{dv}}{{dx}} + v.\dfrac{{du}}{{dx}},\,\,\,\,\,\dfrac{d}{{dx}}\left( {\log A} \right) = \dfrac{1}{A}\dfrac{d}{{dx}}\left( A \right)$ , $\log (A) + \log (B) = \log (AB)$ ${e^{\log (A)}} = A$

Complete step by step solution:
Firstly, writing given problem introducing y on left side as:
y = ${e^{\log \left( {\log x} \right)}}.\log 3x$
Simplifying, exponential term having logarithmic function as in power by using, formula ${e^{\log (A)}} = A$ .
Therefore the term ${e^{\log \left( {\log x} \right)\,}}\,\,becomes\,\,\log x$ .
Then, from above we have
$y = \log x.\log 3x$
Now, differentiating above formed the equation by using the product rule of differentiation on the right hand side. We have,
$\Rightarrow \dfrac{{dy}}{{dx}} = \log 3x\dfrac{d}{{dx}}\left( {\log x} \right) + \log x\dfrac{d}{{dx}}\left( {\log 3x} \right)$
$\dfrac{{dy}}{{dx}} = \log 3x\left( {\dfrac{1}{x}} \right)\dfrac{d}{{dx}}\left( x \right) + \log x\left( {\dfrac{1}{{3x}}} \right)\dfrac{d}{{dx}}\left( {3x} \right)\,\,\,\,\,\,\,(u\sin g\,\,chain\,\,rule\,\,of\,\,differentiation)$
$= \log 3x\left( {\dfrac{1}{x}} \right) \times 1 + \log x\left( {\dfrac{1}{{3x}}} \right) \times 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \dfrac{d}{{dx}}(x) = 1} \right)$
On simplifying right hand side we have,
$= \dfrac{{\log 3x}}{x} + \dfrac{{\log x}}{x}$
Taking the L.C.M. of right hand side of the above equation.
$= \dfrac{1}{x}(\log 3x + \log x)$
= \dfrac{1}{x}\left\\{ {\log \left( {3x} \right)\left( x \right)} \right\\}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\\{ {\because \,\,\,\log A + \log B = \log (AB)} \right\\}
$= \dfrac{{\log \left( {3{x^2}} \right)}}{x}$
Or
$\dfrac{{dy}}{{dx}} = \dfrac{{\log \left( {3{x^2}} \right)}}{x}$
Hence, from above we see that the required derivative of ${e^{\log \left( {\log \,\,x} \right)}}.\,\,\log 3x$ w.r.t. x is $\dfrac{{\log \left( {3{x^2}} \right)}}{x}$ .

Note : While, finding a solution to any math’s problem having exponential term with logarithmic function as a power, never apply a direct formula to simplify it but first to simplify exponential term by using formula and after writing into simpler form proceed with required simplifications.