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A person might prefer to use the logarithmic differentiation for finding derivatives of functions that have another function as an exponent because it is easier to understand and quicker to compute.

Logarithmic Differentiation For Finding Derivatives of Functions with Another Function as the Exponent

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Question: Why might one prefer to use the logarithmic differentiation for finding derivatives of functions that have another function as an exponent? (explain in formal way with points and examples to support your answer)

Asked by noybzzy

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Logarithmic differentiation is a powerful technique for finding the derivatives of functions that have another function as an exponent, or in situations where the regular differentiation process might be cumbersome or difficult to carry out. This method is preferable in such cases due to the following reasons: 1. Simplification of complicated functions: The properties of logarithms, such as the product rule, the quotient rule, and the power rule, allow us to simplify complex functions before differentiating. This can make the process of finding the derivative much cleaner and manageable. For example, when dealing with a function like f(x) = (x^2 + 1)^x, logarithmic differentiation helps in simplifying the exponent. 2. Easy handling of products and quotients: Logarithmic differentiation is specifically useful for differentiating products and quotients of functions. When we take the natural logarithm of a product or quotient, it becomes a sum or difference of logarithms, making it easier to deal with. For instance, if we have a function like g(x) = (x^3 * sin(x))/(e^x * cos(x)), the natural logarithm allows us to isolate each term and differentiate them separately. 3. Tackling variable exponents: Functions with variable exponents can be especially difficult to differentiate using standard methods. By taking the natural logarithm of both sides, we can bring the exponent down, making the function easier to differentiate using chain rule. For example, for a function like h(x) = x^(x^2), logarithmic differentiation assists in handling the exponent (x^2) by bringing it down as a factor. 4. Implicit differentiation: Logarithmic differentiation implicitly allows for the application of implicit differentiation. The natural logarithm creates a relation between the original function and the logarithm, which facilitates the use of implicit differentiation in a more straightforward manner. Example: Let's consider the function y = (x^2 + 1)^x. Using the logarithmic differentiation, we take the natural logarithm of both sides: ln(y) = x * ln(x^2 + 1) Now, we differentiate both sides with respect to x, applying the chain rule and implicit differentiation: (1/y) * dy/dx = (ln(x^2 + 1)) + (x * (2x/(x^2 + 1))) Now we can solve for dy/dx by multiplying both sides by y, which gives us: dy/dx = (x^2 + 1)^x * (ln(x^2 + 1) + (2x^2/(x^2 + 1))) This illustrates that logarithmic differentiation has simplified the differentiation process of a function with another function as an exponent. In conclusion, logarithmic differentiation is a powerful method preferred for handling complicated functions, products, quotients, and variable exponents. By taking the natural logarithm of the given function and applying the properties of logarithms, we can simplify the process of differentiation and obtain the derivative in a less cumbersome way.

Answered by gpt-4

June 3, 2023, 2:51 a.m.

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This article was generated in part by one of OpenAI’s large-scale language-generation model, answering user noybzzy's question. Images were generated with OpenAI's AI model, DALL·E 2. The Internet takes ultimate responsibility for the content of this publication.
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Published: Saturday, June 3, 2023

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