Chapter 4.1, Problem 49E

a.

To determine

To find:the derivative of function in terms of third variable.

Answer to Problem 49E

  $\frac{dy}{dx}=\frac{\cos t}{2t+1}$

Explanation of Solution

Given information:

1. The given equation is as follows

  $x=t^2+t,y=\sin t$ .

2. Need to find the $\frac{dy}{dx}$ .

Formula used:

  $\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$

Calculation:

Give equation of curve is

  $x=t^2+t,y=\sin t$

As $x=t^2+t$ , $\frac{dx}{dt}=\frac{d}{dt}\left(t^2+t\right)$

  $=\frac{d}{dt}{t}^2+\frac{d}{dt}t$

  $=2t+1$

And $y=\sin t$ , $\frac{dy}{dt}=\frac{d}{dt}\sin t$

  $=\cos t$

Calculate $\frac{dy}{dx}$ .

  $\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$

  $=\frac{\cos t}{2t+1}$

Hence $\frac{dy}{dx}=\underset{\_}{\frac{\cos t}{2t+1}}$ .

b.

To determine

To find:the derivative of first derivative with respect to third variable.

Answer to Problem 49E

  $\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\left(\frac{\left(2t+1\right)\sin t+2\cos t}{{\left(2t+1\right)}^2}\right)$

Explanation of Solution

Given information:

1. The calculated first derivative is a follows equation is as follows

  $\frac{dy}{dx}=\frac{\cos t}{2t+1}$ .

2. Need to find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ .

Calculation:

Evaluated equation of $\frac{dy}{dx}$ is as follows

  $\frac{dy}{dx}=\frac{\cos t}{2t+1}$

Differentiating both the sides with respect to $t$ .

  $\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{\cos t}{2t+1}\right)$

  $\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{\cos t}{2t+1}\right)$

  $=\frac{\left(2t+1\right)\frac{d}{dt}\cos t-\cos t\frac{d}{dt}\left(2t+1\right)}{{\left(2t+1\right)}^2}$

  $=\frac{\left(2t+1\right)\left(-\sin t\right)-\cos t.2}{{\left(2t+1\right)}^2}$

  $=\frac{\left(2t+1\right)\left(-\sin t\right)-\cos t.2}{{\left(2t+1\right)}^2}$

  $=-\left(\frac{\left(2t+1\right)\sin t+2\cos t}{{\left(2t+1\right)}^2}\right)$

Hence $\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\left(\frac{\left(2t+1\right)\sin t+2\cos t}{{\left(2t+1\right)}^2}\right)$ .

c.

To determine

To find:the derivative of first derivative.

Answer to Problem 49E

  $\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\left(\frac{\left(2t+1\right)\sin t+2\cos t}{{\left(2t+1\right)}^3}\right)$

Explanation of Solution

Given information:

1. The calculated first derivative is a follows equation is as follows

  $\frac{dy}{dx}=\frac{\cos t}{2t+1}$ .

2. Need to find $\frac{d}{dx}\left(\frac{dy}{dx}\right)$ .

Formula used:

  $\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$

Calculation:

Evaluated equation of $\frac{dy}{dx}$ is as follows

  $\frac{dy}{dx}=\frac{\cos t}{2t+1}$

Differentiating both the sides with respect to $x$ .

  $\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{\cos t}{2t+1}\right)$

  $=\frac{\frac{d}{dt}\left(\frac{\cos t}{2t+1}\right)}{\frac{dx}{dt}}$

  $=\frac{\frac{\left(2t+1\right)\frac{d}{dt}\cos t-\cos t\frac{d}{dt}\left(2t+1\right)}{{\left(2t+1\right)}^2}}{2t+1}$

  $=\frac{\left(2t+1\right)\left(-\sin t\right)-\cos t.2}{{\left(2t+1\right)}^3}$

  $=-\left(\frac{\left(2t+1\right)\sin t+2\cos t}{{\left(2t+1\right)}^3}\right)$

Hence $\frac{d}{dx}\left(\frac{dy}{dx}\right)=-\left(\frac{\left(2t+1\right)\sin t+2\cos t}{{\left(2t+1\right)}^3}\right)$ .

d.

To determine

To find:the expression from above solved parts which represents second dertivative of function with respect to given variable.

Answer to Problem 49E

Part crepresents $\frac{d^{2}y}{d{x}^2}$ .

Explanation of Solution

Given information:

The calculated derivative of first derivative in part c is as follows.

  $\frac{d}{dx}\left(\frac{dy}{dx}\right)=-\left(\frac{\left(2t+1\right)\sin t+2\cos t}{{\left(2t+1\right)}^3}\right)$ .

Calculation:

As equation of $\frac{d^{2}y}{d{x}^2}$ means differentiating $y$ two times with respect to $x$ which is same as differentiating $\frac{dy}{dx}$ once with respect to $x$ that is $\frac{d}{dx}\left(\frac{dy}{dx}\right)$ .

Hence part crepresents $\frac{d^{2}y}{d{x}^2}$ .