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Chapter 4: Problem 6

Differentiate the following functions. \(y=\ln \left(e^{x^{2}+2}\right)\)

Short Answer

Expert verified
\( \frac{dy}{dx} = 2x \)

Step by step solution

01

Simplify the function

Given function is \( y = \ln \left(e^{x^2 + 2}\right) \). Use the logarithm property \( \ln(e^u) = u \) to simplify it. \ y = x^2 + 2 \.
02

Differentiate the simplified function

Differentiate the simplified function \( y = x^2 + 2 \) with respect to \( x \). The derivative of \( x^2 \) is \( 2x \) and the derivative of a constant \( 2 \) is \( 0 \). Therefore, \ \frac{dy}{dx} = 2x \.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

logarithmic differentiation
Logarithmic differentiation is a powerful tool in calculus, especially useful when dealing with complex functions, such as products or quotients of functions, or functions raised to a power. In our exercise, we needed to differentiate a logarithmic function. By understanding the properties of logarithms, we simplified our task significantly. The key property used here is that the natural logarithm of an exponential function simplifies: \( \ln(e^u) = u \). Knowing how to apply this property can save a lot of time and effort in the differentiation process. Using properties of logarithms can turn intimidating expressions into simpler ones, making the differentiation a breeze.
simplifying expressions
Simplifying expressions before differentiating is a crucial step in many calculus problems. In this example, we started with a complicated expression \( y = \ln \(e^{x^2 + 2}\) \). By recognizing that the logarithm and the exponential function are inverse operations, we simplified it to \( y = x^2 + 2 \). This greatly reduced the complexity of the problem. Simplifying expressions helps to avoid mistakes, and it makes the differentiation process much simpler and more straightforward. Remember that recognizing patterns and properties, like those of logarithms and exponentials, is key when simplifying expressions.
derivative rules
Understanding and applying the rules of differentiation is fundamental in calculus. In our simplified expression \( y = x^2 + 2 \), we used both the power rule and the constant rule. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). For \( x^2 \), the derivative is \( 2x \). The constant rule tells us that the derivative of a constant is 0. So, the derivative of 2 is 0. By combining these rules, we quickly find that the derivative of \( y = x^2 + 2 \) is \( \frac{dy}{dx} = 2x \). Mastery of these basic rules allows you to handle more complex differentiation problems with confidence.
calculus
Calculus is the branch of mathematics that studies continuous change. Differentiation, one of the main concepts in calculus, deals with finding the rate at which a function is changing at any given point. This process is crucial in many fields, such as physics, engineering, economics, and beyond. In our exercise, we used several calculus principles to find the derivative of a function. The steps involved understanding the function, simplifying it using logarithmic properties, and then applying derivative rules. Learning calculus equips you with the tools to solve complex problems and understand the changes happening in various natural and man-made systems.

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Most popular questions from this chapter

Simplify the following expressions. \(\frac{3}{2} \ln 4-5 \ln 2\)

Find \(k\) such that \(2^{-x / 5}=e^{k x}\) for all \(x\).

Differentiate the following functions. \(y=\frac{\ln x}{\ln 2 x}\)

Find the equations of the tangent lines to the graph of \(y=\ln |x|\) at \(x=1\) and \(x=-1\).

(a) Find the point on the graph of \(y=e^{-x}\) where the tangent line has slope \(-2\). (b) Plot the graphs of \(y=e^{-x}\) and the tangent line in part (a).
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