/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Differentiate. \(f(x)=\sin x+\... [FREE SOLUTION] | 91Ó°ÊÓ

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Differentiate. \(f(x)=\sin x+\frac{1}{2} \cot x\)

Short Answer

Expert verified
The derivative is \(f'(x) = \cos x - \frac{1}{2} \csc^2 x\).

Step by step solution

01

Differentiate the Sine Function

The derivative of the sine function, \(\sin x\), is the cosine function, \(\cos x\). So the derivative of \(\sin x\) is \(\cos x\).
02

Rewrite the Cotangent Function

Rewrite the \(\cot x\) function as \(\frac{\cos x}{\sin x}\) to make it easier to differentiate.
03

Differentiate the Cotangent Function

Using the quotient rule, differentiate \(\frac{1}{2}\cot x = \frac{1}{2}\frac{\cos x}{\sin x}\). The quotient rule is given by \(\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2}\). Here, \(u = \cos x\) and \(v = \sin x\), so \(u' = -\sin x\) and \(v' = \cos x\). Substitute these into the quotient rule formula:\[\frac{d}{dx}\left(\frac{1}{2}\cot x\right) = \frac{1}{2} \left(\frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{(\sin x)^2}\right)\]
04

Simplify the Quotient Rule Result

Simplify the derivative of \(\frac{1}{2}\cot x\):\[\frac{d}{dx}\left(\frac{1}{2}\cot x\right) = \frac{1}{2} \left(\frac{-\sin^2 x - \cos^2 x}{\sin^2 x}\right) = \frac{1}{2} \left(\frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}\right)\] Since \(\sin^2 x + \cos^2 x = 1\), the expression simplifies further:\[= \frac{1}{2} \left(\frac{-1}{\sin^2 x}\right) = -\frac{1}{2 \sin^2 x}\] which can also be expressed as \(-\frac{1}{2}\csc^2 x\).
05

Combine Derivatives

Combine the derivatives from Step 1 and Step 4:\[f'(x) = \cos x - \frac{1}{2} \csc^2 x\]

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Key Concepts

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

Sine Function Differentiation
Differentiating trigonometric functions is a fundamental skill in calculus, necessary for solving many real-world problems. The sine function, denoted as \( \sin x \), has a derivative that is quite straightforward. When you differentiate \( \sin x \), the result is \( \cos x \). This means that the slope of the tangent line to the curve \( y = \sin x \) at any point \( x \) is given by \( \cos x \).
The reasoning behind this comes down to the limit definition of a derivative. However, for most practical purposes, you can remember this rule as: "The derivative of \( \sin \) is \( \cos \)."
This makes it crucial for tackling various problems that involve periodic phenomena, such as waves or oscillations. As you master this concept, you'll find it easier to differentiate more complex trigonometric functions.
Cotangent Function Differentiation
Moving on to the cotangent function, \( \cot x \) itself is the reciprocal of the tangent function. This makes it slightly trickier to differentiate. It's important to first express \( \cot x \) in terms of sine and cosine: \( \cot x = \frac{\cos x}{\sin x} \).
When differentiating \( \cot x \), we apply the quotient rule because it is a ratio of two functions. The quotient rule states that for a function \( \frac{u(x)}{v(x)} \), its derivative is given by:
  • \( \frac{u'(x)v(x) - u(x)v'(x)}{v^2(x)} \)
For \( \cot x \), \( u = \cos x \) and \( v = \sin x \):
  • The derivative of \( u \), \( u' = -\sin x \)
  • The derivative of \( v \), \( v' = \cos x \)
Plug these into the quotient rule and simplify:
  • \( \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x} \)
  • Using the identity \( \sin^2 x + \cos^2 x = 1 \), this simplifies to \(-\frac{1}{\sin^2 x} \)
This means the derivative of \( \cot x \) is \(-\csc^2 x \). This formula is important when dealing with differentiation tasks involving cotangent.
Quotient Rule
The quotient rule is a powerful differentiation tool, especially when you're faced with a function that is a ratio of two differentiable functions. This rule is particularly useful for differentiating the cotangent function, as we've seen.
The general formula for the quotient rule is:
  • \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \)
To confidently use the quotient rule, follow these steps:
  • Identify \( u \) and \( v \), the numerator and denominator of your function.
  • Find \( u' \), the derivative of \( u \), and \( v' \), the derivative of \( v \).
  • Substitute \( u, v, u', \text{and} v' \) into the quotient rule formula.
With these steps, you can easily manage to differentiate functions like \( \cot x \), leading to results such as \(-\frac{1}{\sin^2 x} \). The quotient rule is also integral in solving complex calculus problems, making it indispensable in your math toolkit.

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Most popular questions from this chapter

Find \(d y / d x\) by implicit differentiation. \(2 x^{3}+x^{2} y-x y^{3}=2\)

(a) Find \(y^{\prime}\) by implicit differentiation. (b) Solve the equation explicitly for \(y\) and differentiate to get \(y^{\prime}\) in terms of \(x .\) (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \(y\) into your solution for part (a). \(x y+2 x+3 x^{2}=4\)

\begin{array}{c}{\text { (a) If } g \text { is differentiable, the Reciprocal Rule says that }} \\ {\frac{d}{d x}\left[\frac{1}{g(x)}\right]=-\frac{g^{\prime}(x)}{[g(x)]^{2}}} \\ {\text { Use the Reciprocal Rule to prove the Reciprocal Rule. }} \\ {\text { (b) Use the Reciprocal Rule to differentiate the function. }}\\\ {y=1 /\left(x^{4}+x^{2}+1\right)} \\ {\text { (c) Use the Reciprocal Rule to verify that the Power Rule }} \\ {\text { is valid for negative integers, that is, }}\\\ \frac{d}{d x}\left(x^{-n}\right)=-n x^{-n-1}\\\ {\text { for all positive integers \(n\). }} \end{array}

Bone mass In Example 1.1 .6 we found an expression for the mass \(m\) of a human femur of length \(L\) in terms of the outer radius \(r,\) the inner radius \(r_{\text { in }},\) and their ratio \(k=r_{\text { in }} / r .\) More generally, if the bone density is \(\rho,\) measured in \(\mathrm{g} / \mathrm{cm}^{3},\) then bone mass is given by the equation \(m=\pi r^{2} L\left[\rho-(\rho-1) k^{2}\right]\) It may happen that both \(\rho\) and \(k\) change with age, \(t\) (a) If \(\rho\) changes during aging, find an expression for the rate of change of \(m\) with respect to \(t\) . (b) If \(k\) changes during aging, find an expression for the rate of change of \(m\) with respect to \(t\) .

(a) Use the definition of the derivative to calculate \(f^{\prime}\) , (b) Check to see that your answer is reasonable by comparing the graphs of \(f\) and \(f^{\prime}\) \(f(t)=t^{2}-\sqrt{t}\)

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