
What is the derivative of $\ln {{x}^{5}}$?
Answer
522.6k+ views
Hint: Assume the given function as $y=f\left( x \right)$. Now, simplify the function by using the property of logarithm given as $\ln {{a}^{m}}=m\ln a$. Now, differentiate both the sides of the assumed function y and use the formula for the derivative of natural log function given as $\dfrac{d\ln x}{dx}=\dfrac{1}{x}$ to get the answer. Use the fact that: if a constant is multiplied to a function then we can directly take that constant out of the derivative.
Complete step by step answer:
Here we have been provided with the function $\ln {{x}^{5}}$ and we are asked to differentiate it. Let us assume this function as y so we have,
$\Rightarrow y=\ln {{x}^{5}}$
Now we need to find the value of $\dfrac{dy}{dx}$. First let us check if we can simplify the given logarithmic function further. Here we have x raised to the power 5 in the argument of log. We know that if we have the argument of log as ‘a’ which is raised to the power ‘m’ then we can simplify it further by using the formula $\ln {{a}^{m}}=m\ln a$. So we can simplify the given logarithmic function as:
$\Rightarrow y=5\ln x$
Differentiating both the sides with respect to x we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( 5\ln x \right)}{dx}$
Here 5 is a constant so it can be taken out of the derivative, so we get,
$\Rightarrow \dfrac{dy}{dx}=5\times \dfrac{d\left( \ln x \right)}{dx}$
Using the formula of the derivative of natural log function given as $\dfrac{d\ln x}{dx}=\dfrac{1}{x}$ we get,
$\begin{align}
& \Rightarrow \dfrac{dy}{dx}=5\times \dfrac{1}{x} \\
& \therefore \dfrac{dy}{dx}=\dfrac{5}{x} \\
\end{align}$
Hence the derivative of the given function is $\dfrac{5}{x}$.
Note: You can also apply the chain rule to solve the question and then simplify. In that case also you will get the same answer however few more will be added. In the chain rule we will first differentiate the function $\ln {{x}^{5}}$ with respect to ${{x}^{5}}$ whose result will be \[\dfrac{1}{{{x}^{5}}}\]. Now, we will differentiate \[{{x}^{5}}\] with respect to x using the formula $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ whose result will be $5{{x}^{4}}$. Finally we will multiply the two derivatives to get the answer. Always remember that for the derivative to exist the function must be defined and for a log function to be defined, its argument must be greater than 0.
Complete step by step answer:
Here we have been provided with the function $\ln {{x}^{5}}$ and we are asked to differentiate it. Let us assume this function as y so we have,
$\Rightarrow y=\ln {{x}^{5}}$
Now we need to find the value of $\dfrac{dy}{dx}$. First let us check if we can simplify the given logarithmic function further. Here we have x raised to the power 5 in the argument of log. We know that if we have the argument of log as ‘a’ which is raised to the power ‘m’ then we can simplify it further by using the formula $\ln {{a}^{m}}=m\ln a$. So we can simplify the given logarithmic function as:
$\Rightarrow y=5\ln x$
Differentiating both the sides with respect to x we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( 5\ln x \right)}{dx}$
Here 5 is a constant so it can be taken out of the derivative, so we get,
$\Rightarrow \dfrac{dy}{dx}=5\times \dfrac{d\left( \ln x \right)}{dx}$
Using the formula of the derivative of natural log function given as $\dfrac{d\ln x}{dx}=\dfrac{1}{x}$ we get,
$\begin{align}
& \Rightarrow \dfrac{dy}{dx}=5\times \dfrac{1}{x} \\
& \therefore \dfrac{dy}{dx}=\dfrac{5}{x} \\
\end{align}$
Hence the derivative of the given function is $\dfrac{5}{x}$.
Note: You can also apply the chain rule to solve the question and then simplify. In that case also you will get the same answer however few more will be added. In the chain rule we will first differentiate the function $\ln {{x}^{5}}$ with respect to ${{x}^{5}}$ whose result will be \[\dfrac{1}{{{x}^{5}}}\]. Now, we will differentiate \[{{x}^{5}}\] with respect to x using the formula $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ whose result will be $5{{x}^{4}}$. Finally we will multiply the two derivatives to get the answer. Always remember that for the derivative to exist the function must be defined and for a log function to be defined, its argument must be greater than 0.
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