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Logarithmic differentiation helps find derivatives of several complicated functions via logarithms. It is also used for functions that involve terms requiring the application of Product Rule or Quotient Rule multiple times for differentiation.
Also Read: Continuity and Differentiability
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Key Terms: Logarithm, Derivatives, Product Rule, Differentiation, Chain Rule, Quotient Rule, Logarithmic value, Initial function
Also Read: Differentiation and Integration Formula
What is Logarithmic Differentiation?
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The method of differentiating functions by first taking its logarithmic value and then differentiating it is called logarithmic differentiation. It is used when differentiating the logarithm of a function is easier than differentiating the function itself. The derivatives become simpler when properties of logarithm and chain rule finding are appropriately used.
| Example: Determine the value of dy/dx, considering that y = \(e^{x^4}\) Solution: As per the given equation, y = \(e^{x^4}\) |
Logarithmic differentiation simplifies the technique by differentiating the logarithm of a function rather than to differentiate the function itself.
Discover about the Chapter video:
Continuity and Differentiability Detailed Video Explanation:
Also Read: Applications of Derivatives
Formula of Logarithmic Differentiation
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Logarithmic Differentiation formula can be given by:
| \(\color{red}{\frac{d}{dx}}\)(xx) = xx (1 + ln x) |
For differentiating, it is vital to consider each side of the given equation. Thus, after assuming log on each side, we get, log y = log [u (x)]{v(x)}
Hence, log y = v(x)log u(x)
Now, applying the chain rule and differentiating either sides with respect to x,
\(\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=v(x) \times \frac{1}{u(x)} \times u^{\prime}x+\log u(x) \times v^{\prime}(x) \\ &\Rightarrow \frac{d y}{d x}=y\left[v(x) \times \frac{1}{u(x)} \times u^{\prime}x+\log u(x) \times v^{\prime}(x)\right] \end{aligned}\)
Usually, the only limitation logarithmic differentiation has is that f(x) and u(x) ought to be positive as logarithmic functions.
Also Read: First Order Differential Equation
Method of Logarithmic Differentiation
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The method to solve Logarithmic Functions include,
- First, consider the natural logarithm of the given function that needs to be differentiated.
- Now, by using the properties of logarithmic functions, distribute the terms which were gathered within the initial function.
- After distribution, differentiate the equation that was acquired as a result.
- Following the same, multiply the equation obtained by the function itself to find the required derivative.
By following the procedure, it becomes simpler to determine the value of logarithmic functions.
Also Read:
Applications of Logarithmic Differentiation
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The applications of logarithmic differentiation include but are not limited to the following.
- Product of Functions: The application of logarithm, for the products of two or more functions, usually changes the product into a sum of functions for simpler differentiation of the function.
- Division of Functions: The differentiation of division of two functions, otherwise known as the Quotient of functions, can be acquired by means of logarithmic differentiation. In simple words, the division of one function with another helps change it into a difference in the logarithm of each of two functions.
- Exponential Function: During exponential functions, two individual functions are considered wherein one is an exponent of another. The application of logarithms to an exponential function is that it changes it into a product of one function, while the exponent of another.
Also Read: Trigonometric Functions
Things to Remember
- The process of differentiating functions by first considering its logarithmic value and then differentiating it subsequently is called logarithmic differentiation.
- The formula used for logarithmic differentiation is \(\frac{d}{dx}\)(xx) = xx (1 + ln x)
- The only limitation logarithmic differentiation has is that f(x) and u(x) should be positive as logarithmic functions while determining an equation.
- The applications of logarithmic differentiation include product of functions, division of functions, and exponential functions.
- By the correct usage of properties of logarithms and chain rule finding, the derivatives become simpler to determine.
Read More: Fundamental Theorem of Calculus
Sample Questions
Ques. Find the derivatives of (cos x)2x. [3 Marks]
Ans. As per the given question, we have, (cos x)2x
Now, after considering log on either side, we get,
Log y = log (cos x)2x
Therefore,
log y = 2x log (cos x)
⇒ y = e2x log cos x
Hence, following the differentiation either side with respect to x, we attain,
\(\frac{dy}{dx}\) = e2x log cos x \(\frac{d}{dx}\) (2x log cos x)
⇒ \(\frac{dy}{dx}\) = (cos x)2x {log cos x. \(\frac{dy}{dx}\)(2x) + 2x . \(\frac{dy}{dx}\) (log cos x)}
⇒ \(\frac{dy}{dx}\) = (cos x)2x {2 log cos x + 2x × \(\frac{1}{\cos x}\) ( – sin x)}
⇒ \(\frac{dy}{dx}\) = (cos x)2x (2 log cos x – 2x tan x)
∴ \(\frac{dy}{dx}\) = 2 (cos x)2x (log cos x – x tan x)
Thus, the derivative of (cos x)2x is 2(cos x)2x (log cos x – x tanx)
Ques. Determine the differentiation of the value xtanx via logarithmic differentiation. [3 Marks]
Ans. As per the equation,
y = xtanx
Now, after applying the logarithms on either side,
log y = log xtanx
log y = tan x × log x.
After differentiating both sides, we get,
\(\frac{d}{dx}\). log y = \(\frac{d}{dx}\) . (tanx . logx)
\(\frac{1}{y}\) . \(\frac{dy}{dx}\) = log x(\(\frac{d}{dx}\) . tanx) + tanx.(\(\frac{d}{dx}\).logx)
\(\frac{1}{y}\) . \(\frac{dy}{dx}\) = log x(sec2x) + tanx(\(\frac{1}{x}\))
\(\frac{1}{y}\) . \(\frac{dy}{dx}\) = sec2x . logx + tanx/x
\(\frac{1}{y}\) . \(\frac{dy}{dx}\) = (x . sec2x . logx + tanx) / x
\(\frac{dy}{dx}\) = y(x. sec2x . logx + tanx) / x
\(\frac{dy}{dx}\) = xtanx(x. sec2x . logx + tanx) / x
Thus, the differentiation of xtanx is xtanx(x. sec2x . logx + tanx) / x
Ques. Determine the value of derivative, considering that y = 2x{cos x}. [2 Marks]
Ans. As per the given equation, y = 2x{cos x}
Now, after applying the logarithms on either side,
Hence, we get, log y = log(2x{cos x})
⇒ log y = log 2 + log xcos x (as log(mn) = log m + log n)
⇒ log y = log 2 + cos x . log x (as log mn = n log m)
Now, differentiating either side via chain rule, we get,
\(\frac{1}{y} \cdot \frac{d y}{d x}=\frac{\cos x}{x}-(\sin x)(\log x)\)
Ques. Determine the derivative of y = xcos (x). [2 Marks]
Ans. As per the equation, y = xcos (x)
Considering y' = xcos (x) . \(\frac{\cos x}{x}\) - sin(x) ln(x)
Assuming ln on each side, ln (y) = g(x) . ln (f(x))
Considering x using product rule,
\(\frac{1}{y}\) . y' = g (x) . \(\frac{f ' (x)}{f(x)}\) + ln(f(x)) . g' (x),
\(y^{\prime}=y\left(g(x) \cdot \frac{f^{\prime}(x)}{f(x)}+\ln (f(x)) \cdot g^{\prime}(x)\right)\)
Thus, it can be said that \(y^{\prime}=f(x)^{g(x)}\left(g(x) \cdot \frac{f^{\prime}(x)}{f(x)}+\ln (f(x)) \cdot g^{\prime}(x)\right)\)
Ques. Determine the Differentiation of: (x + 1)(2x + 3)(5x + 4). [5 Marks]
Ans. As per the question,
y = (x + 1)(2x + 3)(5x + 4)
By using logarithms, Log y = Log (x + 1)(2x + 3)(5x + 4)
And, Logy = Log(x + 1) + Log(2x + 3) + Log(5x + 4)
Thus, \(\frac{d}{dx}\) Logy = \(\frac{d}{dx}\).Log(x + 1) + \(\frac{d}{dx}\).Log(2x + 3) + \(\frac{d}{dx}\).Log(5x + 4)
\(\frac{1}{y}\). \(\frac{dy}{dx}\) = 1/(x + 1).\(\frac{d}{dx}\)(x + 1) + 1/(2x + 3). \(\frac{d}{dx}\)(2x + 3) + 1/(5x + 4).\(\frac{d}{dx}\)(5x + 4)
\(\frac{1}{y}\). \(\frac{dy}{dx}\) = 1/(x + 1).1 + 1/(2x + 3).2 + 1/(5x + 4).5
\(\frac{1}{y}\).\(\frac{dy}{dx}\) = 1/(x + 1) + 2/(2x + 3) + 5/(5x + 4)
\(\frac{dy}{dx}\) = y(1/(x + 1) + 2/(2x + 3) + 5/(5x + 4))
\(\frac{dy}{dx}\) = (x + 1)(2x + 3)(5x + 4)(1/(x + 1) + 2/(2x + 3) + 5/(5x + 4))
\(\frac{dy}{dx}\) = (x + 1)(2x + 3)(5x + 4). 1/(x + 1) + (x + 1)(2x + 3)(5x + 4).2/(2x + 3) + (x + 1)(2x + 3)(5x + 4) .5 / (5x + 4)
\(\frac{dy}{dx}\) = (2x + 3)(5x + 4) + 2(x + 1)(5x + 4) + 5(x + 1)(2x + 3)
\(\frac{dy}{dx}\) = (5x + 4)((2x + 3) + 2(x + 1) + 5(x + 1)(2x + 3)
\(\frac{dy}{dx}\) = (5x + 4)(2x + 3 + 2x + 1) + 5(x + 1)(2x + 3)
\(\frac{dy}{dx}\) = (5x + 4)(4x + 4) + 5(x + 1)(2x + 3)
\(\frac{dy}{dx}\) = 4(5x + 4)(x + 1) + 5(x + 1)(2x + 3)
Therefore, (x + 1) (30x + 31)
Ques. Determine the derivative of y = (x - 1)2 (x - 3)5. [2 Marks]
Ans. Consider the logarithms on either side,
ln y = ln [(x - 1)2 (x - 3)5]
⇒ ln y = ln ((x - 1)2 + ln (x - 3)5
= 2 ln (x - 1) + 5 ln (x - 3)
Now, to determine its logarithmic derivative, we should,
(ln y)' = [2 ln (x - 1) + 5 ln (x - 3)]
⇒ \(\frac{1}{y}\) . y' = 2 . \(\frac{1}{x - 1}\) + 5 . \(\frac{1}{x - 3}\)
⇒ y' = y \(\left( \frac{2}{x - 1} + \frac{5}{x - 3} \right)\)
= (x - 1)2 (x - 3)5 \(\left( \frac{2}{x - 1} + \frac{5}{x - 3} \right)\)
Ques. Establish the differentiation of y = xx. [2 Marks]
Ans. Expression given: y = xx
Here, \(\frac{d}{dx}\) (xn) = nxn-1
⇒ ln y = ln xx
⇒ ln y = x ln x
Now, after differentiation, we will get,
⇒ \(\frac{y^{\prime}}{y}\) = ln x + x \(\left( \frac{1}{x} \right)\)
= ln x + 1
Therefore, y' = y (1 + ln x)
= xx (1 + ln x)
Ques. Determine the second order derivative of log (log x). [2 Marks]
Ans. Considering y = log (log x)
Now, after differentiating with respect to x,
\(\begin{aligned} \frac{d}{d x}&=\frac{d(\log (\log x))}{d x} \\ \frac{d}{d x}&=\frac{1}{\log x} \cdot \frac{d(\log x)}{d x} \\ \frac{d}{d x}&=\frac{1}{\log x} \cdot \frac{1}{x} \\ &=\frac{1}{\log x} \cdot \frac{1}{x} \end{aligned}\)
\(\begin{aligned} &\therefore \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left[(x \log x)^{-1}\right]=(-1) \cdot(x \log x)^{-2} \cdot \frac{d}{d x}(x \log x) \\ &=\frac{-1}{(x \log x)^{2}} \cdot\left[\log x \cdot \frac{d}{d x}(x)+x \frac{d}{d x}(\log x)\right] \\ &=\frac{-1}{(x \log x)^{2}}\left[\log x \cdot 1+x \frac{1}{x}\right]=\frac{-(1+\log x)}{(x \log x)^{2}} \end{aligned}\)
Ques. What is the differentiable value with respect to x of (log x)log x, x > 1. [3 Marks]
Ans. Now, first consider that y = (log x)log x
Now, assuming log on each side,
Log y = log ((log x)log x)
Log y = log x . log (log x)
(It is considered because log (ab) = b log a)
Thus, after differentiating each side,
\(\frac{d(\log y)}{d x}=\frac{d(\log x \cdot \log (\log x))}{d x}\)
\(\begin{aligned} \Rightarrow \frac{d y}{d x}&=y\left[\log (\log x) \frac{1}{x}+\log x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)\right] \\ \Rightarrow \frac{d y}{d x}&=y\left[\frac{1}{x} \log (\log x)+\frac{1}{x}\right] \\ \therefore \frac{d y}{d x}&=(\log x)^{\log x}\left[\frac{1}{x}+\frac{\log (\log x)}{x}\right] \end{aligned}\)
Ques. Differentiate the following term: (log x)x + xlog x. [3 Marks]
Ans. First, consider y = (log x)x + xlog x
Let u = (log x)x, v = xlogx
Therefore,
y = u + v
Now, differentiating each side, we get,
\(\begin{aligned} &\frac{d y}{d x}=\frac{d(u+v)}{d x} \\ &\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \end{aligned}\)
\(\begin{aligned} &\Rightarrow \frac{1}{b} \cdot \frac{d b}{d x}=\frac{1}{x} \cdot x+\log x \\ &\Rightarrow \frac{d b}{d x}=b(1+\log x)=x^{\log x}(1+\log x) \\ &\Rightarrow \frac{d y}{d x}=\frac{d a}{d x}+\frac{d b}{d x}=(\log x)^{x}(1+\log x)+x^{\log x}(1+\log x)) \end{aligned}\)
\(\begin{aligned} &=(1+\log x) \left( (\log x)^{x}+x^{\log x} \right) \\ &=(1+\log x) y \end{aligned}\)
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