Chapter 5: Problem 17
Differentiate the following functions. $$ u=\frac{1+\sin x}{1-\sin x} $$
Short Answer
Expert verified
Answer: The derivative of the function $$u(x)$$ is $$u'(x) = \frac{2 \cos x}{(1-\sin x)^2}$$.
Step by step solution
01
Identify the numerator and denominator functions
First, let's identify the numerator and denominator functions in the given function.
$$
u(x) = \frac{1+\sin x}{1-\sin x}
$$
Here, the numerator function is $$f(x) = 1+\sin x$$ and the denominator function is $$g(x) = 1-\sin x$$.
02
Find the derivatives of the numerator and denominator functions
Next, we need to find the derivatives of the numerator and denominator functions.
For the numerator function:
$$
f(x) = 1+\sin x \Rightarrow f'(x) = 0+\cos x = \cos x
$$
For the denominator function:
$$
g(x) = 1-\sin x \Rightarrow g'(x) = 0-\cos x = -\cos x
$$
03
Apply the quotient rule to find the derivative of u
Now, we'll apply the quotient rule to find the derivative of $$u$$.
$$
u'(x) = \frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2} = \frac{\cos x(1-\sin x) - (-\cos x)(1+\sin x)}{(1-\sin x)^2}
$$
04
Simplify the derivative
Lastly, let's simplify the derivative:
$$
u'(x) = \frac{\cos x - \cos x \sin x + \cos x + \cos x \sin x }{(1-\sin x)^2} = \frac{2 \cos x}{(1-\sin x)^2}
$$
The derivative of the function $$u(x)$$ is:
$$
u'(x) = \frac{2 \cos x}{(1-\sin x)^2}
$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
Differentiating a function that is a quotient of two other functions is a common task in calculus. The quotient rule is a formula used for this purpose and is given by: \[ u'(x) = \frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2} \]
where \( u(x) = f(x)/g(x) \), and \( f(x) \) and \( g(x) \) represent the numerator and the denominator of the function, respectively. It's essential to first find the derivatives of the numerator and the denominator individually before applying this rule. Once applied, simplifying the resulting expression is often necessary for a clear and concise final answer. When simplifying, look for common factors in the numerator and denominator or opportunities to cancel terms out for a more straightforward expression.
where \( u(x) = f(x)/g(x) \), and \( f(x) \) and \( g(x) \) represent the numerator and the denominator of the function, respectively. It's essential to first find the derivatives of the numerator and the denominator individually before applying this rule. Once applied, simplifying the resulting expression is often necessary for a clear and concise final answer. When simplifying, look for common factors in the numerator and denominator or opportunities to cancel terms out for a more straightforward expression.
Differentiation of Trigonometric Functions
Trigonometric functions such as sine and cosine are frequently encountered in calculus, and their derivatives must be memorized for efficient problem solving. The derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \( -\sin x \). Remembering these basic derivatives is crucial when working with more complex calculus problems that involve trigonometric functions. Differentiation can often change the trigonometric function involved, so it's essential to be familiar with the derivatives of all the basic trig functions, including tangent, cotangent, secant, and cosecant, to navigate calculus problems effectively.
Simplifying Derivatives
The process of simplifying derivatives is a crucial step to ensure that the final answer is presented in the simplest form possible. This includes combining like terms, canceling out common factors, and factoring whenever possible. Simplification can reveal the underlying behavior of the derivative more clearly and can sometimes provide insights into critical points, intervals of increase or decrease, and asymptotic behavior. It is also often required for the derivative to be in an appropriate form to apply further calculus concepts, such as finding limits, critical points, and solving optimization problems.
Calculus Problems
Calculus problems often involve applying a sequence of steps and rules to find derivatives, integrals, limits, and solving equations involving functions. A methodical approach ensures that nothing is overlooked. For derivatives, this involves identifying the function type – whether it's a basic function, a quotient, a product, or a composite function – and then applying the corresponding rule: the power rule, the quotient rule, the product rule, or the chain rule. It's vital to work methodically through calculus problems, ensuring each step follows from the previous one, using proper simplification techniques to make subsequent steps easier. Problems should be approached systematically, with each theorem or rule applied carefully and precisely for accurate results.