Math Problem Statement

For the following composite​ function, find an inner function uequals​g(x) and an outer function yequalsf ​(u) such that yequals​f(g(x)). Then calculate StartFraction dy Over dx EndFraction . yequalscosine StartFraction 8 x Over 5 EndFraction Question content area bottom Part 1 Select the correct choice below and fill in the answer box to complete your choice. A. StartFraction dy Over dx EndFraction equalsStartFraction d Over du EndFraction ​(u)timesStartFraction d Over dx EndFraction left parenthesis StartFraction 8 x Over 5 EndFraction right parenthesis equals    enter your response here B. StartFraction dy Over dx EndFraction equalsStartFraction d Over du EndFraction left parenthesis StartFraction 8 u Over 5 EndFraction right parenthesis timesStartFraction d Over dx EndFraction ​(cosine x​)equals    enter your response here C. StartFraction dy Over dx EndFraction equalsStartFraction d Over du EndFraction ​(cosine u​)timesStartFraction d Over dx EndFraction left parenthesis StartFraction x Over 5 EndFraction right parenthesis equals    enter your response here D. StartFraction dy Over dx EndFraction equalsStartFraction d Over du EndFraction ​(cosine u​)timesStartFraction d Over dx EndFraction left parenthesis StartFraction 8 x Over 5 EndFraction right parenthesis equals

Solution

To solve the problem, we need to express the given function as a composite function of two functions, find the derivative of the composite function, and calculate dydx\frac{dy}{dx}.

The given function is: y=cos(8x5)y = \cos\left(\frac{8x}{5}\right)

Step 1: Break it into an inner and outer function

  • Inner function: u=g(x)=8x5u = g(x) = \frac{8x}{5}
  • Outer function: y=f(u)=cos(u)y = f(u) = \cos(u)

This gives the composite function y=f(g(x))=cos(8x5)y = f(g(x)) = \cos\left(\frac{8x}{5}\right).

Step 2: Chain Rule Application

By the chain rule, we calculate the derivative of the composite function: dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

  • dydu\frac{dy}{du} is the derivative of cos(u)\cos(u) with respect to uu, which is: dydu=sin(u)\frac{dy}{du} = -\sin(u)
  • dudx\frac{du}{dx} is the derivative of 8x5\frac{8x}{5} with respect to xx, which is: dudx=85\frac{du}{dx} = \frac{8}{5}

Step 3: Combine the Results

Now, substitute u=8x5u = \frac{8x}{5} back into the expression: dydx=sin(8x5)85\frac{dy}{dx} = -\sin\left(\frac{8x}{5}\right) \cdot \frac{8}{5}

Thus, the derivative is: dydx=85sin(8x5)\frac{dy}{dx} = -\frac{8}{5} \sin\left(\frac{8x}{5}\right)

Answer Selection

The correct choice is: D. dydx=ddu(cos(u))ddx(8x5)=sin(8x5)85\frac{dy}{dx} = \frac{d}{du}(\cos(u)) \cdot \frac{d}{dx}\left(\frac{8x}{5}\right) = -\sin\left(\frac{8x}{5}\right) \cdot \frac{8}{5}


Do you want more details, or have any questions?

Here are five related questions to explore:

  1. How do we apply the chain rule to more complex functions?
  2. What is the derivative of a composite trigonometric function like sin(g(x))\sin(g(x))?
  3. How does the chain rule apply when there are multiple inner functions?
  4. Can we find the second derivative of this composite function?
  5. How would the derivative change if the function were tan(8x5)\tan\left(\frac{8x}{5}\right)?

Tip: Always identify the inner and outer functions clearly when applying the chain rule for composite functions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Composite Functions
Trigonometric Functions
Chain Rule

Formulas

y = cos(8x/5)
dy/dx = (dy/du) * (du/dx)
dy/du = -sin(u)
du/dx = 8/5

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12