MathAdmin: /* Exercise 2: Find equation of tangent line */

2015-11-24T01:45:10Z

Exercise 2: Find equation of tangent line

MathAdmin: Created page with " == Background == So far, you may only have differentiated functions written in the form $y=f(x)$ . But some functions are better descr..."

2015-11-24T01:37:16Z

Created page with " == Background == So far, you may only have differentiated functions written in the form <math style="vertical-align: -5px">y=f(x)</math>. But some functions are better descr..."

New page

== Background ==

So far, you may only have differentiated functions written in the form <math style="vertical-align: -5px">y=f(x)</math>. But some functions are better described by an equation involving <math>x</math> and <math style="vertical-align: -4px">y</math>. For example, <math style="vertical-align: -5px">x^{2}+y^{2}=16</math> describes the graph of a circle with center <math>\left(0,0\right)</math> and radius 4, and is really the graph of two functions <math style="vertical-align: -5px">y=\pm\sqrt{16-x^{2}}</math>, the upper and lower semicircles:

{|cellpadding = "10" align="center"
|[[File:Upper_semicircle.png|400px]]
|
|[[File:Lower_semicircle.png|400px]]
|}

Sometimes, functions described by equations in <math style="vertical-align: 0px">x</math> and <math style="vertical-align: -5px">y</math> are too hard to solve for <math style="vertical-align: -5px">y</math>, for example <math style="vertical-align: -5px">x^{3}+y^{3}=6xy</math>. This equation really describes 3 different functions of x, whose graph is the curve:

{|cellpadding="25" align="center"
|[[File:Curve.png|350px]]
|}
We want to find derivatives of these functions without having to solve for <math style="vertical-align: -5px">y</math> explicitly. We do this by implicit differentiation. The process is to take the derivative of both sides of the given equation with respect to <math>x</math>, and then do some algebra steps to solve for <math style="vertical-align: -5px">y'</math> (or <math>\dfrac{dy}{dx}</math> if you prefer), keeping in mind that <math style="vertical-align: -4px">y</math> is a function of <math style="vertical-align: 0px">x</math> throughout the equation.

== Warm-up exercises ==

Given that <math style="vertical-align: -4px">y</math> is a function of <math style="vertical-align: 0px">x</math>, find the derivative of the
following functions with respect to <math style="vertical-align: 0px">x</math>.

'''1)'''   <math style="vertical-align: -5px">y^{2}</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|<math style="vertical-align: -5px">2yy'</math>
|-
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Reason:  
|-
|Think <math style="vertical-align: -5px">y=f(x)</math>, and view it as <math style="vertical-align: -5px">(f(x))^{2}</math> to see that the derivative is <math style="vertical-align: -5px">2f(x)\cdot f'(x)</math> by the chain rule, but write it as <math style="vertical-align: -4px">2yy'</math>.
|-
|}

'''2)'''   <math style="vertical-align: -5px">xy</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|<math style="vertical-align: -5px">xy'+y</math>
|-
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Reason:  
|-
|<math>x</math> and <math style="vertical-align: -5px">y</math> are both functions of <math>x</math> which are being multiplied together, so the product rule says it's <math style="vertical-align: -5px">x\cdot y'+y\cdot1</math>.
|-
|}

'''3)'''   <math style="vertical-align: -5px">\cos y</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|<math style="vertical-align: -5px">-y'\sin y</math>
|-
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Reason:  
|-
|The function <math style="vertical-align: -5px">y</math> is inside of the cosine function, so the chain rule gives <math style="vertical-align: -5px">(-\sin y)\cdot y'=-y'\sin y</math>.
|-
|}

'''4)'''   <math style="vertical-align: -5px">\sqrt{x+y}</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|<math style="vertical-align: -20px">\frac{1+y'}{2\sqrt{x+y}}</math>
|-
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Reason:  
|-
|Write it as <math style="vertical-align: -5px">(x+y)^{\frac{1}{2}}</math>, and use the chain rule to get   <math style="vertical-align: -15px">\frac{1}{2}\left(x+y\right)^{-\frac{1}{2}}\cdot\left(1+y'\right)</math>, then simplify.
|-
|}

== Exercise 1: Compute ''y''' ==

Find <math style="vertical-align: -4px">y'</math> if <math style="vertical-align: -4px">\sin y-3x^{2}y=8</math>.

Note the <math style="vertical-align: -4px">\sin y</math> term requires the chain rule, the <math style="vertical-align: -4px">3x^{2}y</math>  term needs the product rule, and the derivative of 8 is 0.

We get
::<math>\begin{array}{rcl}
\sin y-3x^{2}y & = & 8\\
\left(\cos y\right)y'-\left(3x^{2}y'+6xy\right) & = & 0\quad (\text{derivative of both sides with respect to } x)\\
\left(\cos y\right)y'-3x^{2}y' & = & 6xy\\
\left(\cos y-3x^{2}\right)y' & = & 6xy\\
y' & = & \dfrac{6xy}{\cos y-3x^{2}}.
\end{array}</math>

== Exercise 2: Find equation of tangent line ==

Find the equation of the tangent line to <math style="vertical-align: -4px">x^{2}+2xy-y^{2}+x=2</math>  at the point <math>\left(1,0\right)</math>.

We first compute <math style="vertical-align: -5px">y'</math> by implicit differentiation.

::<math>\begin{array}{rcl}
x^{2}+2xy-y^{2}+x & = & 2\\
2x+2xy'+2y-2yy'+1 & = & 0\\
x+xy'+y-yy'+\frac{1}{2} & = & 0\\
xy'-yy' & = & -x-y-\frac{1}{2}\\
(x-y)y' & = & -(x+y+\frac{1}{2})\\
y' & = & -\dfrac{x+y+\frac{1}{2}}{x-y}
\end{array}</math>

At the point <math>\left(1,0\right)</math>, we have <math style="vertical-align: -1px">x=1</math> and <math>y=0</math>. Plugging these into our equation for <math style="vertical-align: -5px">y'</math> gives

::<math>\begin{array}{rcl}
y' & = & -\dfrac{1+0+\frac{1}{2}}{1-0} = -\frac{3}{2}.\\

\end{array}</math>

This means the slope of the tangent line at <math>\left(0,1\right)</math> is <math style="vertical-align: -13px">m=-\frac{3}{2}</math>, and a point on this line is <math>\left(1,0\right)</math>. Using the point-slope form of a line, we get

::<math>\begin{array}{rcl}
y-0 & = & -\frac{3}{2}\left(x-1\right)\\
\\
y & = & -\frac{3}{2}x+\frac{3}{2}.\\
\end{array}</math>

Here's a picture of the curve and tangent line:

[[File:Tangent_line_and_curve.png|300px]]

== Exercise 3: Compute ''y"'' ==

Find <math style="vertical-align: -5px">y''</math> if <math>ye^{y}=x</math>.

Use implicit differentiation to find <math style="vertical-align: -5px">y'</math> first:

::<math>\begin{array}{rcl}
ye^{y} & = & x\\
ye^{y}y'+y'e^{y} & = & 1\\
y'\left(ye^{y}+e^{y}\right) & = & 1\\
y' & = & \dfrac{1}{ye^{y}+e^{y}}\\
& = & \left(ye^{y}+e^{y}\right)^{-1}
\end{array}</math>

Now <math style="vertical-align: -5px">y''</math> is just the derivative of <math style="vertical-align: -5px">\left(ye^{y}+e^{y}\right)^{-1}</math> with respect to <math>x</math>. This will require the chain rule. Notice we already found the derivative of <math style="vertical-align: -5px">ye^{y}</math> to be <math style="vertical-align: -5px">ye^{y}y'+y'e^{y}</math>.

So

::<math>\begin{array}{rcl}
y'' & = & -1\left(ye^{y}+e^{y}\right)^{-2}\left(ye^{y}y'+y'e^{y}+e^{y}y'\right)\\
\\
& = & \dfrac{-1}{\left(ye^{y}+e^{y}\right)^{2}}\left(ye^{y}y'+2y'e^{y}\right)\\
\\
& = & -\dfrac{y'e^{y}\left(y+2\right)}{\left(e^{y}\right)^{2}\left(y+1\right)^{2}}\\
\\
& = & -\dfrac{y'\left(y+2\right)}{e^{y}\left(y+1\right)^{2}}\quad (\text{since } e^{y}\neq0).
\end{array}</math>

But we mustn't leave <math style="vertical-align: -5px">y'</math> in our final answer. So, plug <math>y'=\dfrac{1}{e^{y}\left(y+1\right)}</math>
back in to get

::<math>\begin{array}{rcl}
y'' & = & -\dfrac{\frac{1}{e^{y}\left(y+1\right)}\left(y+2\right)}{e^{y}\left(y+1\right)^{2}}\\
\\
& = & -\dfrac{y+2}{\left(e^{y}\right)^{2}\left(y+1\right)^{3}}
\end{array}</math>

as our final answer.

~Page created by Jordan Tousignant

@@ Line 124: / Line 124: @@
 \end{array}</math>
-This means the slope of the tangent line at <math>\left(0,1\right)</math> is <math style="vertical-align: -13px">m=-\frac{3}{2}</math>, and a point on this line is <math>\left(1,0\right)</math>. Using the point-slope form of a line, we get
+This means the slope of the tangent line at <math>\left(1,0\right)</math> is <math style="vertical-align: -13px">m=-\frac{3}{2}</math>, and a point on this line is <math>\left(1,0\right)</math>. Using the point-slope form of a line, we get
 ::<math>\begin{array}{rcl}

Implicit Differentiation - Revision history

MathAdmin: /* Exercise 2: Find equation of tangent line */

MathAdmin: Created page with " == Background == So far, you may only have differentiated functions written in the form y=f(x). But some functions are better descr..."

MathAdmin: Created page with " == Background == So far, you may only have differentiated functions written in the form $y=f(x)$ . But some functions are better descr..."