Question

Find the first derivative and the second derivative of each of the following functions.

Answer 1

To analyze an object's motion, we calculate the first and second derivatives of position to find velocity and acceleration, respectively, and then evaluate them at specific times to understand instant speeds and directional changes.

Step-by-step explanation:

Finding Derivatives and Interpreting Motion

To solve problems related to motion, it is crucial to be able to find the first and second derivatives of position functions, as they correspond to velocity and acceleration respectively. Let's dive into the steps involved in calculating derivatives and analyzing motion:

Find the functional form of the acceleration: This is done by taking the first derivative of the velocity function with respect to time. Since acceleration is the rate of change of velocity, differentiating velocity gives us acceleration.

Find Instantaneous Velocity: To find the instantaneous velocity at a given time (t = 1, 2, 3, and 5 seconds), we directly evaluate the velocity function at these times.

Find Instantaneous Acceleration: Similarly, we find acceleration at specific times by evaluating the acceleration function, which is the derivative of the velocity function, at those times.

Interpret the Results: By examining the instantaneous velocity and acceleration, we can determine the direction of motion and whether the object is speeding up or slowing down. If both the velocity and acceleration vectors are in the same direction, the object is speeding up; if they are in opposite directions, it is slowing down.

By calculating and interpreting these derivatives, we can gain a comprehensive understanding of an object's motion in terms of its speed, velocity, and acceleration at any given instant.

Answer 2

To find the first derivative and second derivative of the following functions.

Now,

a)

`y=4x^7+5x^3`

The first derivative is given by,

`\begin{gathered} y^(\prime)=(d)/(dx)(4x^7+5x^3) \\ =4*7x^(7-1)+5*3x^(3-1) \\ =28x^6+15x^2 \end{gathered}`

The second derivative is,

`\begin{gathered} y^(\prime\prime)=\frac{d^{}}{dx^{}}(28x^6+15x^2) \\ =28*6x^(6-1)+15*2x^(2-1) \\ =168x^5+30x \end{gathered}`

b)

`y=(2)/(x^4)`

The first derivative is,

`\begin{gathered} y^(\prime)=(d)/(dx)((2)/(x^4)) \\ =(d)/(dx)(2x^(-4)) \\ =2*-4x^(-4-1) \\ =-8x^(-5) \\ =-(8)/(x^5) \end{gathered}`

The second derivative is,

`\begin{gathered} y^(\prime\prime)=(d)/(dx)(-(8)/(x^5)) \\ =(d)/(dx)(-8x^(-5)) \\ =-8*-5x^(-5-1) \\ =40x^(-6) \\ =(40)/(x^6) \end{gathered}`

c)

`x=36\sqrt[3]{t}`

The first derivative is,

`\begin{gathered} x^(\prime)=(d)/(dt)(36\sqrt[3]{t}) \\ =(d)/(dt)(36* t^{(1)/(3)}) \\ =36*(1)/(3)* t^{(1)/(3)-1} \\ =12* t^{(1-3)/(3)} \\ =12* t^{-(2)/(3)} \\ =\frac{12}{\sqrt[3]{t^2}} \end{gathered}`

The second derivative is,

`\begin{gathered} x^(\prime\prime)=(d)/(dt)(\frac{12}{\sqrt[3]{t^2}}) \\ =(d)/(dt)(12* t^{-(2)/(3)}) \\ =12*-(2)/(3)* t^{-(2)/(3)-1} \\ =-8t^{-(5)/(3)} \\ =-\frac{8}{t\sqrt[3]{t^2}} \end{gathered}`

d)

`x=4e^(5t+3)`

The first derivative is,

`\begin{gathered} x^(\prime)=(d)/(dt)(4e^(5t+3)) \\ =4* e^(5t+3)*(5+0) \\ =20e^(5t+3) \end{gathered}`

The second derivative is,

`\begin{gathered} x^(\prime\prime)=(d)/(dt)(20e^(5t+3)) \\ =20* e^(5t+3)*(5+0)_{} \\ =100e^(5t+3) \end{gathered}`

e)

`y=16\ln (2x)`

The first derivative is,

`\begin{gathered} y^(\prime)=(d)/(dt)(16\ln (2x)) \\ =16*(1)/(2x)*2 \\ =(16)/(x) \end{gathered}`

The second derivative is,

`\begin{gathered} y^(\prime\prime)=(d)/(dt)((16)/(x)) \\ =(d)/(dt)(16x^(-1)) \\ =16*-1x^(-1-1) \\ =-16x^(-2) \\ =-(16)/(x^2) \end{gathered}`

f)

`x=4\sin (3\theta-1)+5\cos (2\theta+7)`

The first derivate is,

`\begin{gathered} x^(\prime)=(d)/(d\theta)(4\sin (3\theta-1)+5\cos (2\theta+7)) \\ =4*\cos (3\theta-1)*(3-0)+5*-\sin (2\theta+7)*(2+0) \\ =12\cos (3\theta-1)-10\sin (2\theta+7) \end{gathered}`

The second derivative is,

`\begin{gathered} x^(\prime\prime)=(d)/(d\theta)(12\cos (3\theta-1)-10\sin (2\theta+7)) \\ =12*-\sin (3\theta-1)*(3-0)-10*\cos (2\theta+7)*(2+0) \\ =-36\sin (3\theta-1)-20\cos (2\theta+7) \end{gathered}`