Question

If $Y = {(cosx)^{\log x}} + {(logx)^x}$ then, $\dfrac{{dy}}{{dx}} = {(cosx)^{\log x}}\left[ {logx( - tanx) + \dfrac{1}{x}logcosx} \right] + {(logx)^x}\left[ {\dfrac{1}{{\log x}} + log(logx)} \right]$
State true or false.

Answer 1

Hint: Here, we will divide the question into two parts. The first part is ${(cosx)^{\log x}}$ and the second part is ${(logx)^x}$. Let us assume the first part as u, and the second part as v. Now, differentiate those parts with respect to x. That is we will find $\dfrac{{du}}{{dx}}$ and $\dfrac{{dv}}{{dx}}$ . Then to find $\dfrac{{dy}}{{dx}}$, add $\dfrac{{du}}{{dx}}$ and $\dfrac{{dv}}{{dx}}$.
Here, we will use the following formulas:
$\log \left( {{x^y}} \right) = y\log x$
$\dfrac{d}{{dx}}\log u = \dfrac{1}{u}\dfrac{{du}}{{dx}}$
$\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
$\dfrac{d}{{dx}}\log \cos x = \dfrac{1}{{\cos x}}\dfrac{d}{{dx}}\cos x$
$\dfrac{d}{{dx}}\cos x = - \sin x$
$\dfrac{d}{{dx}}x = 1$

Complete step-by-step solution:
In this question, the given expression is
 $ \Rightarrow Y = {(cosx)^{\log x}} + {(logx)^x}$
Now, we will divide the question into two parts.
The first part is ${(cosx)^{\log x}}$.
Let us assume the first part as u.
Therefore,
$ \Rightarrow u = {(cosx)^{\log x}}$
Taking log both sides.
$ \Rightarrow \log u = \log \left[ {{{(cosx)}^{\log x}}} \right]$
We will use the formula $\log \left( {{x^y}} \right) = y\log x$ on the right-hand side.
$ \Rightarrow \log u = \log x(\log cosx)$
Now, differentiate both sides with respect to x.
$ \Rightarrow \dfrac{d}{{dx}}\log u = \dfrac{d}{{dx}}\log x(\log cosx)$
Let us apply the formula $\dfrac{d}{{dx}}\log u = \dfrac{1}{u}\dfrac{{du}}{{dx}}$ on the left-hand side and $\dfrac{d}{{dx}}\log x(\log cosx) = \log cosx\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}(\log cosx)$ on the right-hand side.
Therefore,
$ \Rightarrow \dfrac{1}{u}\dfrac{{du}}{{dx}} = \log cosx\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}(\log cosx)$
Here, apply the formulas $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$, and $\dfrac{d}{{dx}}\log \cos x = \dfrac{1}{{\cos x}}\dfrac{d}{{dx}}\cos x$.
$ \Rightarrow \dfrac{1}{u}\dfrac{{du}}{{dx}} = \log cosx\left( {\dfrac{1}{x}} \right) + \log x\left( {\dfrac{1}{{\cos x}}} \right)\dfrac{d}{{dx}}cosx$
Now, as we already know that $\dfrac{d}{{dx}}\cos x = - \sin x$
$ \Rightarrow \dfrac{1}{u}\dfrac{{du}}{{dx}} = \dfrac{{\log cosx}}{x} + \dfrac{{\log x}}{{\cos x}}\left( { - \sin x} \right)$
We know that $\dfrac{{\sin x}}{{\cos x}} = \tan x$.
 Therefore,
$ \Rightarrow \dfrac{1}{u}\dfrac{{du}}{{dx}} = \dfrac{{\log cosx}}{x} + \log x\left( { - \tan x} \right)$
Now, multiply both sides by u.
$ \Rightarrow \dfrac{{du}}{{dx}} = u\left( {\dfrac{{\log cosx}}{x} + \log x\left( { - \tan x} \right)} \right)$
Let us substitute the value of u.$u = {(cosx)^{\log x}}$
So,
$ \Rightarrow \dfrac{{du}}{{dx}} = {(cosx)^{\log x}}\left( {\dfrac{{\log cosx}}{x} + \log x\left( { - \tan x} \right)} \right)$
We can also write the above term as below.
$ \Rightarrow \dfrac{{du}}{{dx}} = {(cosx)^{\log x}}\left( {\log x\left( { - \tan x} \right) + \dfrac{1}{x}\log cosx} \right)$..................(1)
Now, let us take the second part.
The second part is ${(logx)^x}$.
Let us assume the first part as v.
Therefore,
$ \Rightarrow v = {(logx)^x}$
Taking log both sides.
$ \Rightarrow \log v = \log \left[ {{{(logx)}^x}} \right]$
We will use the formula $\log \left( {{x^y}} \right) = y\log x$ on the right-hand side.
$ \Rightarrow \log v = x\log (\log x)$
Now, differentiate both sides with respect to x.
$ \Rightarrow \dfrac{d}{{dx}}\log v = \dfrac{d}{{dx}}\left( {x\log (\log x)} \right)$
Let us apply the formula $\dfrac{d}{{dx}}\log v = \dfrac{1}{v}\dfrac{{dv}}{{dx}}$ on the left-hand side and $\dfrac{d}{{dx}}\left( {x\log (\log x)} \right) = \log (\log x)\dfrac{d}{{dx}}x + x\dfrac{d}{{dx}}(\log \left( {\log x} \right))$ on the right-hand side.
Therefore,
$ \Rightarrow \dfrac{1}{v}\dfrac{{dv}}{{dx}} = \log (\log x)\dfrac{d}{{dx}}x + x\dfrac{d}{{dx}}(\log \left( {\log x} \right))$
Here, $\dfrac{d}{{dx}}x = 1$ and $\dfrac{d}{{dx}}(\log \left( {\log x} \right)) = \dfrac{1}{{\log x}}\dfrac{d}{{dx}}\log x$
$ \Rightarrow \dfrac{1}{v}\dfrac{{dv}}{{dx}} = \log (\log x)\left( 1 \right) + x\dfrac{1}{{\log x}}\dfrac{d}{{dx}}\log x$
Apply the formula $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
So,$ \Rightarrow \dfrac{1}{v}\dfrac{{dv}}{{dx}} = \log (\log x) + x\dfrac{1}{{\log x\left( x \right)}}$
That is equal to,
$ \Rightarrow \dfrac{1}{v}\dfrac{{dv}}{{dx}} = \log (\log x) + \dfrac{1}{{\log x}}$
Let us multiply both sides by v.
$ \Rightarrow \dfrac{{dv}}{{dx}} = v\left( {\log (\log x) + \dfrac{1}{{\log x}}} \right)$
Here, substitute$v = {(logx)^x}$.
$ \Rightarrow \dfrac{{dv}}{{dx}} = {(logx)^x}\left( {\log (\log x) + \dfrac{1}{{\log x}}} \right)$
We can also write the above equation.
$ \Rightarrow \dfrac{{dv}}{{dx}} = {(logx)^x}\left( {\dfrac{1}{{\log x}} + \log (\log x)} \right)$................(2)
Now,
$\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}} + \dfrac{{dv}}{{dx}}$
Substitute the values from equations (1) and (2).
$ \Rightarrow \dfrac{{dy}}{{dx}} = {(cosx)^{\log x}}\left( {\log x\left( { - \tan x} \right) + \dfrac{1}{x}\log cosx} \right) + {(logx)^x}\left( {\log (\log x) + \dfrac{1}{{\log x}}} \right)$
Hence, the answer is true.

Note: Logarithmic differentiation: It is a method to find the derivatives of some complicated functions, using logarithms. There are cases in which differentiating the logarithm of a given function is simpler as compared to differentiating the function itself. By proper usage of properties of logarithms and chain rule finding, the derivatives become easy.