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Find the derivative of the function. $$ y=\ln \left|\frac{-1+\sin x}{2+\sin x}\right| $$

Short Answer

Expert verified
The derivative of the function \( y = \ln \left| \frac{-1+\sin x}{2 + \sin x} \right| \) is obtained. It's different for different intervals: For \( \sin x > -1 \), it's \( \frac{ 3\cos x}{-1 + \sin x} \times \frac{1}{(2 + \sin x)^2} \) , and for \( \sin x < -1 \), it's \( -\frac{ 3\cos x}{-1 + \sin x} \times \frac{1}{(2 + \sin x)^2} \).

Step by step solution

01

Write down the given function and consider the modulus function

The function is given as \( y = \ln \left| \frac{-1+\sin x}{2 + \sin x} \right| \). The modulus function changes sign at \(\sin x = -1\) and \(\sin x = -2\). So we need to separately consider three intervals: (-\infty, -2], (-2, -1) and [-1, \infty), and find the derivative accordingly.
02

Assume y = u and Find the derivative of y with respect to \(\sin(x)\) for each interval.

Let us denote \( u = \frac{-1+\sin x}{2 + \sin x} \). For \( \sin x > -1 \), \( y = \ln (u) \), and for \( \sin x < -1 \), \( y = \ln (-u) \).Differentiate with respect to \(\sin x\):The derivative for \( \sin x > -1 \) is \( \frac{1}{u} \frac{du}{d\sin x} \) and for \( \sin x < -1 \) it is \( -\frac{1}{u} \frac{du}{d\sin x} \).
03

Differentiate u with respect to \(\sin x\) and Substitute in the above equations.

Differentiate u, with \(\sin x > -1\) and \(\sin x < -1\), \( \frac{du}{d\sin x} = \frac{(2 + \sin x) - (-1 + \sin x)}{(2 + \sin x)^2} = \frac{3}{(2 + \sin x)^2}\).Substitute \( \frac{du}{d\sin x} \) into the equations from the previous step, we get the derivative for \( \sin x > -1 \): \( \frac{1}{u} \times \frac{3}{(2 + \sin x)^2} \) and \(\sin x < -1\): \( -\frac{1}{u} \times \frac{3}{(2 + \sin x)^2} \).
04

Replace u with the original expression and differentiate \(\sin x\)

Replace u with \(\frac{-1 + \sin x}{2 + \sin x}\). The derivative of \( \sin x \) with respect to x is \( \cos x \). On multiplying by \( \frac{d \sin x}{dx} \), we get the derivative for \( \sin x > -1 \): \( \frac{ \cos x}{(\frac{-1 + \sin x}{2 + \sin x})} \times \frac{3}{(2 + \sin x)^2} \) and for \( \sin x < -1 \): \( -\frac{ \cos x}{(\frac{-1 + \sin x}{2 + \sin x})} \times \frac{3}{(2 + \sin x)^2} \).
05

Simplify the obtained expression

Simplify the obtained expression to find the final derivative. For the interval \( \sin x > -1 \), the derivative will be \( \frac{ 3\cos x}{-1 + \sin x} \times \frac{1}{(2 + \sin x)^2} \) , for \( \sin x < -1 \), the derivative will be \( -\frac{ 3\cos x}{-1 + \sin x} \times \frac{1}{(2 + \sin x)^2} \).

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Key Concepts

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

Derivative of Logarithmic Functions
Logarithmic functions often appear in various mathematical problems, and understanding how to differentiate them is crucial. The derivative of a logarithmic function like \( y = \ln{|u|} \) requires knowing how the expression inside the log function fluctuates.

When dealing with \( y = \ln{|f(x)|} \), the derivative can be found using the chain rule and recognizing that it varies based on the sign of \( f(x) \):
  • For positive \( f(x) \), the derivative is \( \frac{1}{f(x)} \cdot f'(x) \)
  • For negative \( f(x) \), the chain rule results in \( -\frac{1}{f(x)} \cdot f'(x) \)
In the given exercise, the expression \( \frac{-1 + \sin x}{2 + \sin x} \) was treated in intervals to handle positive and negative outcomes separately. Differentiating logarithmic functions in this manner ensures that the absolute value sign is respected, accurately reflecting the nature of the expression inside the logarithm.
Differentiation Techniques
Differentiating a complex function requires precise techniques, especially when dealing with compositional and multi-variable scenarios. Differentiation techniques such as the chain rule, quotient rule, and product rule are essential tools.

For functions like \( y = \ln|f(x)| \), we employ the chain rule extensively. The chain rule suggests taking the derivative of the outer function (the logarithm) and multiplying by the derivative of the inner function (\( f(x) \)).
In the exercise, we considered:\[ y = \ln\left|\frac{-1 + \sin x}{2 + \sin x}\right| \]The function \( u = \frac{-1 + \sin x}{2 + \sin x} \) inside the logarithm required careful attention. By differentiating \( u \) with respect to \( \sin x \) and then multiplying by the derivative of \( \sin x \) with respect to \( x \), you ensure all components are neatly handled.
  • First, differentiate \( u \), using \( \frac{du}{d\sin x} \)
  • Then use the derivative of \( \sin x \), which is \( \cos x \)
Combining these derivatives accurately allows you to seamlessly apply the chain rule for the overall expression, exemplifying a versatile differentiation approach.
Trigonometric Functions Differentiation
In calculus, differentiation of trigonometric functions is a fundamental concept. These functions, such as \( \sin x \), \( \cos x \), and others, often appear within more complex expressions and require specific understanding of their behavior.

For the function \( y = \ln\left|\frac{-1 + \sin x}{2 + \sin x}\right| \), \( \sin x \) was central in its differentiation. Here's what you need about differentiating such scenarios:
  • The derivative of \( \sin x \) is \( \cos x \)
  • Understand that \( \sin x \) transforms various parts of your equation, influencing the resulting derivative
  • The behavior of trigonometric functions can dictate intervals of function behavior
During the solution, acknowledging where \( \sin x > -1 \) and \( \sin x < -1 \), allows you to correctly allocate where the derivative expressions change, reflecting the properties of \( \sin x \). Such differentiation not only aids in solving complex problems but also deepens your grasp of trigonometric identities and their calculus relationships.

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Most popular questions from this chapter

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

Find the integral. $$ \int \frac{\operatorname{csch}(1 / x) \operatorname{coth}(1 / x)}{x^{2}} d x $$

Graph \(y_{1}=\frac{x}{1+x^{2}}, y_{2}=\arctan x\), and \(y_{3}=x\) on \([0,10]\). Prove that \(\frac{x}{1+x^{2}}<\arctan x0\).

Find \(\left(f^{-1}\right)^{\prime}(a)\) for the function \(f\) and the given real number \(a\). \(f(x)=\sqrt{x-4}, \quad a=2\)

Find the derivative of the function. $$ y=2 x \sinh ^{-1}(2 x)-\sqrt{1+4 x^{2}} $$

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