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Question: How do you differentiate \(y=\ln x\)?...

How do you differentiate y=ln⁡xy=\ln x?

Explanation

Solution

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⇔ey=xy=\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 xx and apply the formulas ddx(f(g(x)))=f′(g(x))×g′(x)\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right), ddx(x)=1\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⁡xy=\ln x
We can observe natural logarithmic function in the given equation. From the logarithmic ruley=ln⁡x⇔ey=xy=\ln x\Leftrightarrow {{e}^{y}}=x we can write the above equation as
ey=x.....(i){{e}^{y}}=x.....\left( \text{i} \right)
Now the given equation is converted in exponential form.
Differentiating the above equation with respect to xx, then we will get
ddx(ey)=ddx(x)\dfrac{d}{dx}\left( {{e}^{y}} \right)=\dfrac{d}{dx}\left( x \right)
Using the formula ddx(f(g(x)))=f′(g(x))×g′(x)\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
ddx(ey)×ddx(y)=ddx(x)\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 ddx(ey)=ey\dfrac{d}{dx}\left( {{e}^{y}} \right)={{e}^{y}}. Substituting this value in the above equation, then we will get
ey×dydx=ddx(x){{e}^{y}}\times \dfrac{dy}{dx}=\dfrac{d}{dx}\left( x \right)
We know that the value ddx(x)=1\dfrac{d}{dx}\left( x \right)=1, so the above equation is modified as
eydydx=1{{e}^{y}}\dfrac{dy}{dx}=1
From equation (i)\left( \text{i} \right) we have the value ey=x{{e}^{y}}=x. Substituting this value in the above equation, then we will get
xdydx=1x\dfrac{dy}{dx}=1
Dividing the above equation with xx on both sides, then we will have
dydx=1x\dfrac{dy}{dx}=\dfrac{1}{x}

Hence the derivative of the given equation y=ln⁡xy=\ln x is 1x\dfrac{1}{x}.

Note: We can also solve this problem by using the basic definition of the derivative which is given by ddx(f(x))=lim⁡δx→0f(x+δx)−f(x)δx\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.