Question

Find the derivative of the following function. $y=\left| \dfrac{{{\log }_{3}}{{x}^{-5}}}{{{\log }_{3}}243}-1 \right|-5$

Answer 1

- Hint: To find the derivative of the function given in the question, one must start by simplifying the given function using properties of logarithmic function and then differentiating the terms given in the function using sum and product rule of differentiation.

Complete step-by-step solution -

To find the derivative of the function $y=\left| \dfrac{{{\log }_{3}}{{x}^{-5}}}{{{\log }_{3}}243}-1 \right|-5$, we will differentiate it with respect to the variable x using some logarithmic properties.
We will first simplify the given function.
We know that ${{\log }_{b}}a=\dfrac{\log a}{\log b}$.
Substituting $a={{x}^{-5}},b=3$, we get${{\log }_{3}}{{x}^{-5}}=\dfrac{\log {{x}^{-5}}}{\log 3}.....\left( 1 \right)$.
We know that $\log {{a}^{b}}=b\log a$.
Substituting $a=x,b=-5$, we get $\log {{x}^{-5}}=-5\log x$.
Substituting the above equation in equation (1), we get ${{\log }_{3}}{{x}^{-5}}=\dfrac{\log {{x}^{-5}}}{\log 3}=\dfrac{-5\log x}{\log 3}.....\left( 2 \right)$.
We know that 243 can be factorized as $243={{3}^{5}}$.
We know that ${{\log }_{a}}{{a}^{b}}=b{{\log }_{a}}a=b$.
Substituting $a=3,b=5$, we get ${{\log }_{3}}243={{\log }_{3}}{{3}^{5}}=5.....\left( 3 \right)$.
Substituting equation (2) and (3) in the given equation of function, we get $y=\left| \dfrac{{{\log }_{3}}{{x}^{-5}}}{{{\log }_{3}}243}-1 \right|-5=\left| \dfrac{-5\log x}{\log 3}\times \dfrac{1}{5}-1 \right|-5=\left| \dfrac{-\log x}{\log 3}-1 \right|-5.....\left( 4 \right)$.
Case1: If $x>1$, we have $\log x>0$.Thus, we have $y=\left| \dfrac{-\log x}{\log 3}-1 \right|-5=\dfrac{\log x}{\log 3}+1-5=\dfrac{\log x}{\log 3}-4$.
We know that differentiation of any function of the form $y=a\log x+b$ is $\dfrac{dy}{dx}=\dfrac{a}{x}$.
Substituting $a=\dfrac{1}{\log 3},b=-4$ in the above equation, we have $\dfrac{dy}{dx}=\dfrac{1}{x\log 3}$.
We know that the differentiation of a constant function with respect to any variable is 0.
Thus, differentiation of the function $y=\left| \dfrac{{{\log }_{3}}{{x}^{-5}}}{{{\log }_{3}}243}-1 \right|-5$ is $\dfrac{dy}{dx}=\dfrac{1}{x\log 3}$ if $x>1$.
Case2: If $x<1$, we have $\log x<0$.Thus, we have $\dfrac{-\log x}{\log 3}>0$. We will remove the modulus depending if $\dfrac{-\log x}{\log 3}$ is greater or less than 1.
Case2 (a): If $\dfrac{-\log x}{\log 3}>1$, we have $y=\left| \dfrac{-\log x}{\log 3}-1 \right|-5=\dfrac{-\log x}{\log 3}-1-5=\dfrac{-\log x}{\log 3}-6$.
We know that differentiation of any function of the form $y=a\log x+b$ is $\dfrac{dy}{dx}=\dfrac{a}{x}$.
Substituting $a=\dfrac{-1}{\log 3},b=-6$ in the above equation, we have $\dfrac{dy}{dx}=\dfrac{-1}{x\log 3}$.
We know that the differentiation of a constant function with respect to any variable is 0.
Thus, differentiation of the function $y=\left| \dfrac{{{\log }_{3}}{{x}^{-5}}}{{{\log }_{3}}243}-1 \right|-5$ is $\dfrac{dy}{dx}=\dfrac{-1}{x\log 3}$ if $x<1$ and $\dfrac{-\log x}{\log 3}>1$.
Case2 (b): If $\dfrac{-\log x}{\log 3}<1$, we have $y=\left| \dfrac{-\log x}{\log 3}-1 \right|-5=\dfrac{\log x}{\log 3}+1-5=\dfrac{\log x}{\log 3}-4$.
We know that differentiation of any function of the form $y=a\log x+b$ is $\dfrac{dy}{dx}=\dfrac{a}{x}$.
Substituting $a=\dfrac{1}{\log 3},b=-4$ in the above equation, we have $\dfrac{dy}{dx}=\dfrac{1}{x\log 3}$.
We know that the differentiation of a constant function with respect to any variable is 0.
Thus, differentiation of the function $y=\left| \dfrac{{{\log }_{3}}{{x}^{-5}}}{{{\log }_{3}}243}-1 \right|-5$ is $\dfrac{dy}{dx}=\dfrac{1}{x\log 3}$ if$x<1$ and $\dfrac{-\log x}{\log 3}<1$.

Note: The first derivative of any function signifies the slope of the function. Also, we get different values of derivatives of the function based on different values of x. Thus, one should remove modulus carefully considering all the cases.