Math Problem Statement

Kerjakan

Kerjakan turunan pertama, turunan implisit, kecepatan benda, aturan rantai, dan turunan kedua dari fungsi yang diberikan.

Solution

Let's go through the problems one by one.

1. Derivatives of f(x)=x43x3+2x25x+7f(x) = x^4 - 3x^3 + 2x^2 - 5x + 7

To calculate the first derivative f(x)f'(x): f(x)=ddx(x43x3+2x25x+7)f'(x) = \frac{d}{dx} \left(x^4 - 3x^3 + 2x^2 - 5x + 7\right) f(x)=4x39x2+4x5f'(x) = 4x^3 - 9x^2 + 4x - 5

2. Implicit Differentiation of x2+y2=25x^2 + y^2 = 25

To find dydx\frac{dy}{dx}, differentiate both sides with respect to xx: ddx(x2+y2)=ddx(25)\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25) 2x+2ydydx=02x + 2y \frac{dy}{dx} = 0 dydx=xy\frac{dy}{dx} = -\frac{x}{y}

3. Velocity from Position Function s(t)=3t36t2+9t4s(t) = 3t^3 - 6t^2 + 9t - 4

To find the velocity, we need to differentiate the position function s(t)s(t): v(t)=dsdt=ddt(3t36t2+9t4)v(t) = \frac{ds}{dt} = \frac{d}{dt} \left( 3t^3 - 6t^2 + 9t - 4 \right) v(t)=9t212t+9v(t) = 9t^2 - 12t + 9 At t=2t = 2: v(2)=9(2)212(2)+9=3624+9=21v(2) = 9(2)^2 - 12(2) + 9 = 36 - 24 + 9 = 21 Thus, the velocity at t=2t = 2 is 21m/s21 \, \text{m/s}.

4. Chain Rule: f(x)=(3x24x+5)7f(x) = (3x^2 - 4x + 5)^7

Using the chain rule, let u=3x24x+5u = 3x^2 - 4x + 5, so: f(x)=7u6dudxf'(x) = 7u^6 \cdot \frac{du}{dx} First, calculate dudx\frac{du}{dx}: dudx=6x4\frac{du}{dx} = 6x - 4 Now substitute back: f(x)=7(3x24x+5)6(6x4)f'(x) = 7(3x^2 - 4x + 5)^6 \cdot (6x - 4)

5. Second Derivative of f(x)=sin(3x)f(x) = \sin(3x)

The first derivative: f(x)=ddxsin(3x)=3cos(3x)f'(x) = \frac{d}{dx} \sin(3x) = 3 \cos(3x) Now, the second derivative: f(x)=ddx(3cos(3x))=9sin(3x)f''(x) = \frac{d}{dx} (3 \cos(3x)) = -9 \sin(3x) The second derivative describes the acceleration or change in the rate of the angular velocity in the context of rotational motion.


Do you want more details on any specific step or explanation?

Here are 5 related questions for practice:

  1. What is the second derivative of f(x)=x43x3+2x25x+7f(x) = x^4 - 3x^3 + 2x^2 - 5x + 7?
  2. If x2+y2=36x^2 + y^2 = 36, what is dydx\frac{dy}{dx} in terms of xx and yy?
  3. What is the acceleration a(t)=dvdta(t) = \frac{dv}{dt} for the position function s(t)=3t36t2+9t4s(t) = 3t^3 - 6t^2 + 9t - 4?
  4. Find the derivative of f(x)=(2x2+3x1)5f(x) = (2x^2 + 3x - 1)^5 using the chain rule.
  5. What is the third derivative of f(x)=sin(3x)f(x) = \sin(3x)?

Tip: When working with implicit differentiation, remember to apply the chain rule when differentiating terms involving yy, as yy is a function of xx.

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Math Problem Analysis

Mathematical Concepts

Derivatives
Implicit Differentiation
Velocity from Position Function
Chain Rule
Second Derivative

Formulas

f'(x) = 4x^3 - 9x^2 + 4x - 5
Implicit Differentiation: dy/dx = -x/y
Velocity: v(t) = ds/dt = 9t^2 - 12t + 9
Chain Rule: f'(x) = 7(3x^2 - 4x + 5)^6 * (6x - 4)
Second Derivative: f''(x) = -9 sin(3x)

Theorems

Power Rule for Derivatives
Implicit Differentiation
Chain Rule

Suitable Grade Level

University Level (Calculus I or II)