Question

How do you find the derivative of $y={{\left( \ln x \right)}^{n}}$ ? \[\]

Answer 1

Hint: We recall the definition of composite function $gof\left( x \right)=g\left( f\left( x \right) \right)$. We recall the chain rule of differentiation $\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$ where $y=gof={{\left( \ln x \right)}^{n}}$ and $u=f\left( x \right)=\ln x$. We first find $u=f\left( x \right)$ as the function inside the bracket and $y$ as the given function and then differentiate using chain rule. We then solve alternatively using the first principle as $\dfrac{d}{dx}f\left( x \right)=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$.\[\]

Complete step-by-step answer:
If the functions $f\left( x \right),g\left( x \right)$ are defined within sets $f:A\to B$ and $g:B\to C$ then the composite function from A to C is defend as $g\left( f\left( x \right) \right)$ within sets $gof:A\to C$. If we denote $g\left( f\left( x \right) \right)=y$ and $f\left( x \right)=u$ then we can differentiate the composite function using chain rule as
\[\dfrac{d}{dx}g\left( f\left( x \right) \right)=\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}\]
We are asked to differentiate the function ${{\left( \ln x \right)}^{n}}$. We see that it is a composite function which made by functions polynomial function that is ${{x}^{n}}$ and natural logarithmic function that is $\ln x$. Let us assign the function within the bracket as $f\left( x \right)=\ln x=u$ and $g\left( x \right)={{x}^{n}}$. So we have $g\left( f\left( x \right) \right)=g\left( \ln x \right)={{\left( \ln x \right)}^{n}}=y$. We differentiate using chain rule to have;
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx} \\
 & \Rightarrow \dfrac{d}{dx}y=\dfrac{d}{du}y\times \dfrac{d}{dx}u \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=\dfrac{d}{d\left( \ln x \right)}{{\left( \ln x \right)}^{n}}\times \dfrac{d}{dx}\left( \ln x \right) \\
\end{align}\]
We know that from standard differentiation of polynomial function as $\dfrac{d}{dt}{{t}^{n}}=n{{t}^{n-1}}$ where $n$ is any real number. We use it for $t=\ln x,n=n$ in the above step to have
\[\Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=n{{\left( \ln x \right)}^{n-1}}\times \dfrac{d}{dx}\left( \ln x \right)\]
We know that from standard differentiation of logarithmic function $\dfrac{d}{dt}\ln t=\dfrac{1}{t}$. We use it for $t=x$ in the above step to have
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=n{{\left( \ln x \right)}^{n-1}}\times \dfrac{1}{x} \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=\dfrac{n{{\left( \ln x \right)}^{n-1}}}{x} \\
\end{align}\]

Alternative Method: We can find the derivative using the first principle. We know that derivative using the first principle for any continuous function $f\left( x \right)$ at any point in the domain of $f$ is obtained from the following working rule
\[\dfrac{d}{dx}f\left( x \right)=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}\]
We put $f\left( x \right)={{\left( \ln x \right)}^{n}}$ in the working rule as
\[\dfrac{d}{dx}\ln {{\left( x+h \right)}^{n}}=\displaystyle \lim_{h\to 0}\dfrac{{{\left( \ln \left( x+h \right) \right)}^{n}}-{{\left( \ln x \right)}^{n}}}{h}.....\left( 1 \right)\]
We use the algebraic identity ${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}+{{a}^{n-1}}b+{{a}^{n-1}}{{b}^{2}}+...+a{{b}^{n-2}}+{{b}^{n-1}} \right)$ and simplify the numerator of the above step for $a=\ln \left( x+h \right),b=\ln x$ to have
\[\begin{align}
  & {{\left( \ln \left( x+h \right) \right)}^{n}}-{{\left( \ln x \right)}^{n}}=\left( \ln \left( x+h \right)-\ln x \right)\left\{ {{\left( \ln \left( x+h \right) \right)}^{n-1}}+{{\left( \ln \left( x+h \right) \right)}^{n-2}}\left( \ln x \right)+ \right. \\
 & \left. ...+\left( \ln \left( x+h \right) \right){{\left( \ln x \right)}^{n-2}}+{{\left( \ln x \right)}^{n-1}} \right\} \\
\end{align}\]
We also know the logarithmic identity involving quotient $\log \left( \dfrac{m}{n} \right)=\log m-\log n$ for $m=x+h,n=x$ in the above step to have;
\[\begin{align}
  & {{\left( \ln \left( x+h \right) \right)}^{n}}-{{\left( \ln x \right)}^{n}}=\ln \left( \dfrac{x+h}{x} \right)\left\{ {{\left( \ln \left( x+h \right) \right)}^{n-1}}+{{\left( \ln \left( x+h \right) \right)}^{n-2}}\left( \ln x \right)+ \right. \\
 & \left. ...+\left( \ln \left( x+h \right) \right){{\left( \ln x \right)}^{n-2}}+{{\left( \ln x \right)}^{n-1}} \right\} \\
\end{align}\]
We put the above result in (1) and then use law of product of limits to have;
\[\begin{align}
  & \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=\displaystyle \lim_{h\to 0}\dfrac{{{\left( \ln \left( x+h \right) \right)}^{n}}-{{\left( \ln x \right)}^{n}}}{h} \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=\displaystyle \lim_{h\to 0}\dfrac{\ln \left( \dfrac{x+h}{x} \right)}{h}\times \displaystyle \lim_{h\to 0}\left\{ {{\left( \ln \left( x+h \right) \right)}^{n-1}}+{{\left( \ln \left( x+h \right) \right)}^{n-2}}\left( \ln x \right)+ \right. \\
 & \left. ...+\left( \ln \left( x+h \right) \right){{\left( \ln x \right)}^{n-2}}+{{\left( \ln x \right)}^{n-1}} \right\} \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=\displaystyle \lim_{h\to 0}\dfrac{\ln \left( \dfrac{x}{x}+\dfrac{h}{x} \right)}{h}\times \left\{ {{\left( \ln x \right)}^{n-1}}+{{\left( \ln x \right)}^{n-1}}+....\left( n\text{ times} \right) \right\} \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=\displaystyle \lim_{h\to 0}\dfrac{\ln \left( 1+\dfrac{h}{x} \right)}{h}\times n{{\left( \ln x \right)}^{n-1}} \\
\end{align}\]
We multiply and divide $\dfrac{1}{x}$ in the numerator and denominator of the limit to have
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=\displaystyle \lim_{h\to 0}\dfrac{\dfrac{1}{x}\ln \left( 1+\dfrac{h}{x} \right)}{\dfrac{1}{x}\times h}\times n{{\left( \ln x \right)}^{n-1}} \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=n{{\left( \ln x \right)}^{n-1}}\times \dfrac{1}{x}\times \displaystyle \lim_{h\to 0}\dfrac{\ln \left( 1+\dfrac{h}{x} \right)}{\dfrac{h}{x}} \\
\end{align}\]
We use the standard limit $\displaystyle \lim_{t\to 0}\dfrac{\ln \left( 1+t \right)}{t}=1$ for $t=\dfrac{h}{x}$ in the above step to have;
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=n{{\left( \ln x \right)}^{n-1}}\times \dfrac{1}{x}\times 1 \\
 & \Rightarrow \dfrac{d}{dx}{{\left( \ln x \right)}^{n}}=\dfrac{n{{\left( \ln x \right)}^{n-1}}}{x} \\
\end{align}\]

Note: We note that logarithmic function takes only positive real numbers as inputs and hence the given function ${{\left( \ln x \right)}^{n}}$ will have domain and range as ${{\left( \ln x \right)}^{n}}:{{\mathsf{\mathbb{R}}}^{+}}\to \mathsf{\mathbb{R}}$. We also note that while we are finding the limit in first derivative principle the terms independent of $h$ are constants like $\dfrac{1}{x},n,{{\left( \ln x \right)}^{n-1}}$since we are taking limit on $h$ not on $x$. If limit exists for $f\left( x \right),g\left( x \right)$ at $x=a$ then by law of product of limits $\displaystyle \lim_{x \to a}f\left( x \right)g\left( x \right)=\displaystyle \lim_{x \to a}f\left( x \right)\cdot \displaystyle \lim_{x \to a}g\left( x \right)$.