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Chapter 3: Problem 24

Use Version 2 of the Chain Rule to calculate the derivatives of the following composite functions. $$y=\cos (5 t+1)$$

Short Answer

Expert verified
Answer: The derivative of the function y = cos(5t + 1) is dy/dt = -5sin(5t + 1).

Step by step solution

01

Find the derivative of f(u) with respect to u

The function $$f(u) = \cos(u)$$. To find its derivative, we use the formula: $$\frac{d}{du} \cos(u) = -\sin(u)$$ So, $$f'(u) = -\sin(u)$$.
02

Find the derivative of g(t) with respect to t

The function $$g(t) = 5t + 1$$. To find its derivative, we use the formula: $$\frac{d}{dt} (5t + 1) = 5$$ So, $$g'(t) = 5$$.
03

Apply the Chain Rule

Now, let's use the Chain Rule to find the derivative of the composite function y: $$\frac{dy}{dt} = f'(g(t)) \cdot g'(t)$$ Plugging in the values of f'(u) and g'(t) that we calculated in Steps 1 and 2: $$\frac{dy}{dt} = (-\sin(g(t))) \cdot (5)$$ Now, let's substitute $$g(t)$$ into the equation: $$\frac{dy}{dt} = (-\sin(5t + 1)) \cdot (5)$$ Therefore, the derivative of the composite function $$y=\cos(5t + 1)$$ is: $$\frac{dy}{dt} = -5\sin(5t + 1)$$

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Key Concepts

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

Composite Functions
Composite functions involve combining two or more functions, creating a new one. Here, we have two functions:
  • The outer function, which is \(f(u) = \cos(u)\)
  • The inner function, \(g(t) = 5t + 1\)
These functions are combined such that the output of the inner function becomes the input of the outer function. This form is often referred to as \'nested functions\'. For example, if each part were calculated step by step:
  • Calculate \(g(t)\) to get \(u = 5t + 1\)
  • Then, plug \(u\) into \(f\),\ which is \(\cos(u)\)
  • The result becomes \(y = \cos(5t + 1)\)
Understanding composite functions is important because they appear frequently in calculus and everyday mathematics. A clear grasp of them is essential for successful derivative calculations, especially when using the Chain Rule.
Derivative Calculation
Calculating derivatives is a fundamental operation in calculus. The derivatives provide information about the rate at which a function is changing at any given point. In the context of composite functions, derivative calculation involves applying the Chain Rule.
The Chain Rule is a formula for calculating the derivative of a composite function. It states that if you have a composite function \(y = f(g(t))\), then the derivative is given by:\[\frac{dy}{dt} = f'(g(t)) \cdot g'(t)\]Let's break this down:
  • First, find the derivative of the outer function \(f(u)\) with respect to its input \(u\). This gives \(f'(u)\).
  • Next, find the derivative of the inner function \(g(t)\) with respect to \(t\), giving \(g'(t)\).
  • Finally, multiply these results to find the derivative of the composite function \(\frac{dy}{dt}\).
In our example, this process results in:
  • Outer function: \(f'(g(t)) = -\sin(g(t))\);
  • Inner function: \(g'(t) = 5\);
  • Composite function derivative: \(\frac{dy}{dt} = -5\sin(5t + 1)\).
Trigonometric Functions
Trigonometric functions, like \(\cos(u)\), are vital in many areas of mathematics and applied sciences, including calculus. They describe the relationships of the angles and sides of triangles, specifically in a unit circle.
For this exercise, you need to consider the derivative property of the cosine function, which is uniquely characterized among trigonometric derivatives. The derivative of \(\cos(u)\) is \(-\sin(u)\).\
This means:
  • Whenever you differentiate \(\cos(u)\), the result is \(-\sin(u)\).
  • This derivative indicates how the rate of change of the \(\cos(u)\) function behaves over time.
In problems involving composite functions, the derivative of trigonometric functions can be more complex. Using the Chain Rule, derivatives of trigonometric functions contribute to computing the overall change rate in the composite function.

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Most popular questions from this chapter

Vertical tangent lines a. Determine the points at which the curve \(x+y^{3}-y=1\) has a vertical tangent line (see Exercise 52 ). b. Does the curve have any horizontal tangent lines? Explain.

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals. . A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Use implicit differentiation if needed to find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=c x^{2} ; x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants

Derivatives and inverse functions Suppose the slope of the curve \(y=f^{-1}(x)\) at (4,7) is \(\frac{4}{5}\) Find \(f^{\prime}(7)\)

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}\left[\frac{f(x)}{(x+2)}\right]\right|_{x=4}$$

Graphing \(f\) and \(f^{\prime}\) a. Graph \(f\) with a graphing utility. b. Compute and graph \(f^{\prime}\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has \(a\) horizontal tangent line. $$f(x)=(x-1) \sin ^{-1} x \text { on }[-1,1]$$
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