Math Problem Statement

0:07:54 Question 1 Not yet answered Marked out of 1.00 Let f(x)=x^2 and g(x)=x^3-4 then the derivative of f(g(x)) will be Question 1Select one:

a. 2(x^3+4) * 3x

b. (3x)^2 * 2(x^3-4)

c. (3x^2) * 2(x^3-4)

d. 3x^2

e. 2(x^3-4)

Question 2 Not yet answered Marked out of 1.00 The derivative when using the Chain Rule is the Question 2Select one:

a. derivative of the outside leaving the inside changed times the derivative of the inside.

b. derivative of the outside leaving the outside unchanged times the derivative of the inside.

c. derivative of the outside leaving the inside unchanged times the derivative of the inside.

d. derivative of the inside leaving the outside unchanged times the derivative of the outside.

e. none of these

Question 3 Not yet answered Marked out of 1.00 The derivative of y = (5/3) * ln(2-3x) by using the chain rule is Question 3Select one:

a. -15/(6-9x)

b. 5/3 * (-3)

c. none of these

d. -1/(6-9x)

e. 15/(6+9x)

Question 4 Not yet answered Marked out of 1.00 The slope of a parametrized curve is given by: Question 4Select one:

a. dy/dt = (dy/dt) / (dx/dt)

b. dt/dx = (dy/dt) / (dx/dt)

c. none of these

d. dy/dx = (dy/dt) / (dx/dt)

e. dy/dx = (dx/dt) / (dy/dt)

Question 5 Not yet answered Marked out of 1.00 A curve is defined by the parametric equations x = (3t + 2)^2 and y = 2t^3 + 9. Question 5Select one:

a. dy/dt = (t^2)/(3t+2)

b. dy/dx = (2t)/(3t-2)

c. dy/dx = (t^2)/(3t)

d. dy/dx = (t^2)/(3t+2)

e. dt/dx = (t^2)/(3t+2)

Solution

Let's tackle each question one by one.

Question 1:

We are asked to find the derivative of f(g(x))f(g(x)) where:

  • f(x)=x2f(x) = x^2
  • g(x)=x34g(x) = x^3 - 4

By applying the chain rule, the derivative of f(g(x))f(g(x)) is: ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)

  • f(x)=x2f(x) = x^2, so f(x)=2xf'(x) = 2x
  • g(x)=x34g(x) = x^3 - 4, so g(x)=3x2g'(x) = 3x^2

Thus: ddx[f(g(x))]=2g(x)g(x)=2(x34)3x2\frac{d}{dx} [f(g(x))] = 2g(x) \cdot g'(x) = 2(x^3 - 4) \cdot 3x^2

Therefore, the correct answer is c. (3x2)2(x34)(3x^2) \cdot 2(x^3 - 4).

Question 2:

The chain rule states that to differentiate a composite function, you:

  • Take the derivative of the outside function, leaving the inside function unchanged, then
  • Multiply by the derivative of the inside function.

The correct answer is c. "derivative of the outside leaving the inside unchanged times the derivative of the inside."

Question 3:

We are asked to find the derivative of y=53ln(23x)y = \frac{5}{3} \ln(2 - 3x).

Using the chain rule:

  • The derivative of ln(u)\ln(u) is 1u\frac{1}{u}, so: ddxln(23x)=123x(3)\frac{d}{dx} \ln(2 - 3x) = \frac{1}{2 - 3x} \cdot (-3) The 3-3 comes from the derivative of the inside function 23x2 - 3x.

Thus, the full derivative is: ddx(53ln(23x))=53323x=523x\frac{d}{dx} \left( \frac{5}{3} \ln(2 - 3x) \right) = \frac{5}{3} \cdot \frac{-3}{2 - 3x} = \frac{-5}{2 - 3x}

Since this doesn't match any of the options exactly, the correct answer is c. "none of these."

Question 4:

The derivative of a parametrized curve is given by: dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} Thus, the correct answer is d. dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

Question 5:

We are given the parametric equations:

  • x=(3t+2)2x = (3t + 2)^2
  • y=2t3+9y = 2t^3 + 9

First, let's compute the derivatives dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}:

  • dxdt=2(3t+2)3=6(3t+2)\frac{dx}{dt} = 2(3t + 2) \cdot 3 = 6(3t + 2)
  • dydt=6t2\frac{dy}{dt} = 6t^2

Now, using the chain rule: dydx=dy/dtdx/dt=6t26(3t+2)=t23t+2\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t^2}{6(3t + 2)} = \frac{t^2}{3t + 2}

Thus, the correct answer is d. dydx=t23t+2\frac{dy}{dx} = \frac{t^2}{3t + 2}.


Let me know if you'd like further details or explanations!

Here are 5 related questions you might find useful:

  1. How do you apply the chain rule to more complex composite functions?
  2. What is the product rule, and how does it differ from the chain rule?
  3. How do you find the second derivative of a parametric curve?
  4. Can you differentiate a parametric equation when xx and yy are both functions of more than one variable?
  5. How do you differentiate implicit functions?

Tip: Always simplify inside functions first when using the chain rule—it makes the calculations more manageable.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Chain Rule
Parametric Equations

Formulas

Chain rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)
Derivative of logarithmic functions: d/dx[ln(u)] = 1/u * du/dx
Parametric derivative: dy/dx = (dy/dt) / (dx/dt)

Theorems

Chain Rule
Logarithmic Differentiation

Suitable Grade Level

Undergraduate Calculus or AP Calculus