
How do you differentiate $y=\ln x$?
Answer
531k+ views
Hint: In this problem we need to calculate the derivative of the given function. We can observe that the given function has natural logarithmic function. So, we can use the logarithmic rule which is $y=\ln x\Leftrightarrow {{e}^{y}}=x$. From this we can convert the given logarithmic function into exponential function. Now we will differentiate the obtained equation with respect to $x$ and apply the formulas $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$, $\dfrac{d}{dx}\left( x \right)=1$. Now we will simplify the obtained equation by using mathematical operations to get the required result.
Complete step by step answer:
Given equation is $y=\ln x$
We can observe natural logarithmic function in the given equation. From the logarithmic rule$y=\ln x\Leftrightarrow {{e}^{y}}=x$ we can write the above equation as
${{e}^{y}}=x.....\left( \text{i} \right)$
Now the given equation is converted in exponential form.
Differentiating the above equation with respect to $x$, then we will get
$\dfrac{d}{dx}\left( {{e}^{y}} \right)=\dfrac{d}{dx}\left( x \right)$
Using the formula $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$ in the above equation, then we will have
$\dfrac{d}{dx}\left( {{e}^{y}} \right)\times \dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}\left( x \right)$
We have the differentiation formula $\dfrac{d}{dx}\left( {{e}^{y}} \right)={{e}^{y}}$. Substituting this value in the above equation, then we will get
${{e}^{y}}\times \dfrac{dy}{dx}=\dfrac{d}{dx}\left( x \right)$
We know that the value $\dfrac{d}{dx}\left( x \right)=1$, so the above equation is modified as
${{e}^{y}}\dfrac{dy}{dx}=1$
From equation $\left( \text{i} \right)$ we have the value ${{e}^{y}}=x$. Substituting this value in the above equation, then we will get
$x\dfrac{dy}{dx}=1$
Dividing the above equation with $x$ on both sides, then we will have
$\dfrac{dy}{dx}=\dfrac{1}{x}$
Hence the derivative of the given equation $y=\ln x$ is $\dfrac{1}{x}$.
Note: We can also solve this problem by using the basic definition of the derivative which is given by $\dfrac{d}{dx}\left( f\left( x \right) \right)=\displaystyle \lim_{\delta x \to 0}\dfrac{f\left( x+\delta x \right)-f\left( x \right)}{\delta x}$. If you want to solve this problem by using this formula, we need to apply a lot of formulas and calculations. So, we have not followed this method.
Complete step by step answer:
Given equation is $y=\ln x$
We can observe natural logarithmic function in the given equation. From the logarithmic rule$y=\ln x\Leftrightarrow {{e}^{y}}=x$ we can write the above equation as
${{e}^{y}}=x.....\left( \text{i} \right)$
Now the given equation is converted in exponential form.
Differentiating the above equation with respect to $x$, then we will get
$\dfrac{d}{dx}\left( {{e}^{y}} \right)=\dfrac{d}{dx}\left( x \right)$
Using the formula $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$ in the above equation, then we will have
$\dfrac{d}{dx}\left( {{e}^{y}} \right)\times \dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}\left( x \right)$
We have the differentiation formula $\dfrac{d}{dx}\left( {{e}^{y}} \right)={{e}^{y}}$. Substituting this value in the above equation, then we will get
${{e}^{y}}\times \dfrac{dy}{dx}=\dfrac{d}{dx}\left( x \right)$
We know that the value $\dfrac{d}{dx}\left( x \right)=1$, so the above equation is modified as
${{e}^{y}}\dfrac{dy}{dx}=1$
From equation $\left( \text{i} \right)$ we have the value ${{e}^{y}}=x$. Substituting this value in the above equation, then we will get
$x\dfrac{dy}{dx}=1$
Dividing the above equation with $x$ on both sides, then we will have
$\dfrac{dy}{dx}=\dfrac{1}{x}$
Hence the derivative of the given equation $y=\ln x$ is $\dfrac{1}{x}$.
Note: We can also solve this problem by using the basic definition of the derivative which is given by $\dfrac{d}{dx}\left( f\left( x \right) \right)=\displaystyle \lim_{\delta x \to 0}\dfrac{f\left( x+\delta x \right)-f\left( x \right)}{\delta x}$. If you want to solve this problem by using this formula, we need to apply a lot of formulas and calculations. So, we have not followed this method.
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