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Chapter 3: Problem 4

Find \(d y / d x\). $$y=\sqrt{x} \sec x+3$$

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \sec x + \sqrt{x} \sec x \tan x \)

Step by step solution

01

Identify the Sum of Functions

The given function is a sum: \[ y = \sqrt{x} \sec x + 3 \] which is the sum of \( \sqrt{x} \sec x \) and the constant 3.
02

Differentiate Constant Term

The derivative of a constant, such as 3, is 0. Therefore, the derivative of 3 with respect to \( x \) is 0.
03

Apply Product Rule to \( \sqrt{x} \sec x \)

We have a product of two functions: \( u(x) = \sqrt{x} \) and \( v(x) = \sec x \). The product rule is \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \).
04

Differentiate \( u(x) = \sqrt{x} \)

The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \). Therefore, \( u'(x) = \frac{1}{2\sqrt{x}} \).
05

Differentiate \( v(x) = \sec x \)

The derivative of \( \sec x \) is \( \sec x \tan x \). Thus, \( v'(x) = \sec x \tan x \).
06

Combine Results Using Product Rule

Substitute into the product rule: \[ u'(x)v(x) + u(x)v'(x) = \frac{1}{2\sqrt{x}} \sec x + \sqrt{x} \sec x \tan x \]Combine these results to get the derivative of the product.
07

Finalize the Derivative

Combine terms from the previous steps: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \sec x + \sqrt{x} \sec x \tan x \]This is the derivative of the original function \( y \).

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Key Concepts

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

Product Rule
The product rule is a fundamental concept in calculus that helps us differentiate products of two functions.
It's necessary because differentiating two functions separately and then multiplying them won't work.
The correct approach is given by the formula:
  • \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]
To understand better, consider the function \( y = \sqrt{x} \sec x \).
Here,
  • \( u(x) = \sqrt{x} \) with derivative \( u'(x) = \frac{1}{2\sqrt{x}} \)
  • \( v(x) = \sec x \) with derivative \( v'(x) = \sec x \tan x \)
Utilizing the product rule, you substitute:
  • \( u'(x)v(x) + u(x)v'(x) = \frac{1}{2\sqrt{x}} \sec x + \sqrt{x} \sec x \tan x \)
This results in the derivative of \( y \).
Always remember: Product rule helps tackle multiplication of differentiable functions.
Chain Rule
The chain rule is essential when dealing with compositions of functions in differentiation.
Imagine a nested function structure like \( y = f(g(x)) \).
Here, we need to differentiate the outer function \( f \) and the inner function \( g \) simultaneously.The chain rule formula is:
  • \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)
This means you first find the derivative of the outer function and then multiply by the derivative of the inner function.
In the given example \( y = \sqrt{x} \sec x \), if approached with composite functions,
  • \( \sqrt{x} \) could be thought of as \( x^{1/2} \) - its derivative being straightforward like a chain
Though not directly applied here, understanding chain rule is crucial for more complex compositions encountered in calculus.
Trigonometric Derivatives
Differentiating trigonometric functions is a critical skill in calculus.
Each trig function has its specific rules.
Knowing these rules is essential to correctly handle such derivatives.For instance, in our function \( y = \sqrt{x} \sec x \):
  • The derivative of \( \sec x \) is \( \sec x \tan x \), which you use in our product rule.
This arises from the identity \( \sec x = \frac{1}{\cos x} \) and involves both \( \sec x \) and \( \tan x \).Such exact derivatives for trigonometric functions allow us to craft precise formulas during differentiation.
Make sure to memorize and practice derivatives for all basic trigonometric functions, as these will recur in many problems.
Derivative of a Constant
In calculus, a crucial thing to remember is that the derivative of a constant is always zero.
Whenever you differentiate a constant, it vanishes.In our exercise, the function includes a constant, \( 3 \).
Regardless of its value, when differentiating:
  • \( \frac{d}{dx}[3] = 0 \)
This simplification is beneficial as it helps to focus solely on the variable parts of a function.
Understanding that constants "disappear" simplifies differentiation, especially in polynomial expressions with constant terms.
This becomes intuitive as constants do not change, thus having no impact on the variable increments in calculus.

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Most popular questions from this chapter

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x^{5}+y^{3} x+y x^{2}+y^{4}=4, \quad P(1,1)$$

Find the tangent to \(y=((x-1) /(x+1))^{2}\) at \(x=0\)

Graph the function \(y=2 \cos 2 x\) for \(-2 \leq x \leq 3.5 .\) Then, on the same screen, graph $$y=\frac{\sin 2(x+h)-\sin 2 x}{h}$$ for \(h=1.0,0.5,\) and \(0.2 .\) Experiment with other values of \(h\) including negative values. What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.

Suppose that \(f(x)=x^{2}\) and \(g(x)=|x| .\) Then the Suppose that \(f(x)=x^{2}\) and \(g(x)=|x| .\) Then the composites $$(f \circ g)(x)=|x|^{2}=x^{2} \quad \text { and } \quad(g \circ f)(x)=\left|x^{2}\right|=x^{2}$$ are both differentiable at \(x=0\) even though \(g\) itself is not differentiable at \(x=0 .\) Does this contradict the Chain Rule? Explain.

When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g.\) By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\). a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{s}^{2}\) to a new location. This increases the period by \(d T=0.001\) s. Find \(d g\) and estimate the value of \(g\) at the new location.
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