*Notes to a video lecture on http://www.unizor.com*

__Derivatives - Problems 4__

**Problem 4.1**For an implicitly defined function

*,*

**x=x(t)**

**y=y(t)**find the second derivative

*=*

**y''**_{xx}(t)*d²*.

**y/**d**x**²*Solution*

First derivative

*d*was discussed in the lecture on differentiation of implicitly defined function and is

**y/**d**x***=*

**y'**_{x}(t)*d*= [

**y(t)/**d**x(t)***d*]

**y(t)/**d**t***/*[

*d*]

**x(t)/**d**t**The second derivative is a derivative of the first derivative:

*=*

**y''**_{xx}(t)*d/d*[

**x***] =*

**y'**_{x}(t)=

*d/d*{[

**x***d*]

**y(t)/**d**t***/*[

*d*]} =

**x(t)/**d**t**=

*d/d*[

**x***] =*

**y'**_{t}(t)/x'_{t}(t)= {

*d/d*[

**t***]}*

**y'**_{t}(t)/x'_{t}(t)*/*[

*d*]=

**x(t)**/d**t**= {

*d/d*[

**t***]}*

**y'**_{t}(t)/x'_{t}(t)*/*

*=*

**x'**_{t}(t)= [

*]*

**y''**_{tt}(t)·x'_{t}(t)−y'_{t}(t)·x''_{t}(t)*/*{[

*]²·*

**x'**_{t}(t)*} =*

**x'**_{t}(t)= [

*]*

**y''**_{tt}(t)·x'_{t}(t)−y'_{t}(t)·x''_{t}(t)*/*[

*]³ =*

**x'**_{t}(t)

**Problem 4.2**Using the results of Problem 4.1, find the second derivative of parametrically defined function

*,*

**x=arctan(t)**

**y=ln(1+t²)***Answer*

**y''**_{xx}(t) = 2+2·t²

**Problem 4.3**Using the results of Problem 4.1, find the second derivative of parametrically defined function

*,*

**x=cos(2t)**

**y=sin²(t)***Answer*

**y''**_{xx}(t) = 0This prompts that

**as a function of**

*y**must be a straight line.*

**x**Indeed, since

*,*

**x=cos(2t)=1−2sin²(t)**it follows that

*and*

**x+2y=1***.*

**y = (1−x)/2**Graph of this function is below

The parametric definition of this function puts restrictions on its domain

**−1 ≤ x ≤ 1**and range

*.*

**0 ≤ y ≤ 1**But within these boundaries

*is a linear function.*

**y(x)**

**Problem 4.4**Using the results of Problem 4.1, find the second derivative of parametrically defined function

*,*

**x=cos(2t)**

**y=t²***Answer*

*[*

**y''**_{xx}(t) =*]*

**sin(2t)−2t·cos(2t)***[*

**/***]*

**2sin³(2t)**
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