Implicit Differentiation - Revision history

Grad at 01:59, 24 November 2015

2015-11-24T01:59:47Z

Grad: /* Background */

2015-11-21T23:29:57Z

Background

Grad: /* Exercise 1: Compute y' */

2015-11-21T22:24:07Z

Exercise 1: Compute y'

Grad: /* Warm-up exercises */

2015-11-21T22:22:40Z

Warm-up exercises

Grad: /* Background */

2015-11-21T22:18:36Z

Background

Grad: /* Exercise 3: Compute y" */

2015-11-21T21:36:45Z

Exercise 3: Compute y"

Grad: /* Exercise 3: Compute y" */

2015-11-21T21:35:46Z

Exercise 3: Compute y"

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

2015-11-21T21:31:31Z

Exercise 2: Find equation of tangent line

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

2015-11-21T21:24:10Z

Exercise 2: Find equation of tangent line

Grad: /* Exercise 1: Compute y' */

2015-11-21T20:46:00Z

Exercise 1: Compute y'

@@ 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}

@@ Line 4: / Line 4: @@
 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 = "50" align="center"
+{|cellpadding = "10" align="center"
-|[[File:Upper_semicircle.png|300px]]
+|[[File:Upper_semicircle.png|400px]]
+|
-|[[File:Lower_semicircle.png|300px]]
+|[[File:Lower_semicircle.png|400px]]
 |}
@@ Line 13: / Line 13: @@
 {|cellpadding="25" align="center"
-|[[File:Curve.png|250px]]
+|[[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.

@@ Line 88: / Line 88: @@
 == Exercise 1: Compute ''y''' ==
-Find <math style="vertical-align: -5px">y'</math> if <math style="vertical-align: -5px">\sin y-3x^{2}y=8</math>.
+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: -5px">\sin y</math> term requires the chain rule, the <math style="vertical-align: -5px">3x^{2}y</math>&thinsp; term needs the product rule, and the derivative of 8 is 0.
+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>&thinsp; term needs the product rule, and the derivative of 8 is 0.
 We get

@@ Line 19: / Line 19: @@
 == Warm-up exercises ==
-Given that <math style="vertical-align: -5px">y</math> is a function of <math>x</math>, find the derivative of the
+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>x</math>.
+following functions with respect to <math style="vertical-align: 0px">x</math>.
 '''1)''' &nbsp; <math style="vertical-align: -5px">y^{2}</math>
@@ Line 34: / Line 34: @@
 !Reason: &nbsp;
 |-
-|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: -5px">2yy'</math>.
+|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>.
 |-
 |}

@@ Line 15: / Line 15: @@
 |[[File:Curve.png|250px]]
 |}
-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: -5px">y</math> is a function of <math>x</math> throughout the equation.
+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 ==

@@ Line 162: / Line 162: @@
   & = & -\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)
+  & = & -\dfrac{y'\left(y+2\right)}{e^{y}\left(y+1\right)^{2}}\quad (\text{since } e^{y}\neq0).
 \end{array}</math>

@@ Line 123: / Line 123: @@
 \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
@@ Line 133: / Line 132: @@
 \end{array}</math>
-Here's a picture of the curve and its tangent line:
+Here's a picture of the curve and tangent line:
-[[File:Tangent_line.png]]
+[[File:Tangent_line_and_curve.png|300px]]
 == Exercise 3: Compute ''y"'' ==

@@ Line 103: / Line 103: @@
 == Exercise 2: Find equation of tangent line ==
-Find the equation of the tangent line to <math style="vertical-align: -13px">\tan y=\dfrac{y}{x}</math>&thinsp; at the point &thinsp; <math>\left(\frac{\pi}{4},\frac{\pi}{4}\right)</math>.
+Find the equation of the tangent line to <math style="vertical-align: -4px">x^{2}+2xy-y^{2}+x=2</math>&thinsp; at the point <math>\left(1,0\right)</math>.
-We first compute <math style="vertical-align: -5px">y'</math> by implicit differentiation. Note the derivative of the right side requires the quotient rule.
+We first compute <math style="vertical-align: -5px">y'</math> by implicit differentiation.
 ::<math>\begin{array}{rcl}
-\tan y & = & \dfrac{y}{x}\\
+x^{2}+2xy-y^{2}+x & = & 2\\
-\left(\sec^{2}y\right)y' & = & \dfrac{xy'-y}{x^{2}}\\
+x+2xy'+2y-2yy'+1 & = & 0\\
-x^{2}y'\sec^{2}y & = & xy'-y\\
+x+xy'+y-yy'+\frac{1}{2} & = & 0\\
-x^{2}y'\sec^{2}y-xy' & = & -y\\
+xy'-yy' & = & -x-y-\frac{1}{2}\\
-y'\left(x^{2}\sec^{2}y-x\right) & = & -y\\
+(x-y)y' & = & -(x+y+\frac{1}{2})\\
-y' & = & \dfrac{-y}{x^{2}\sec^{2}y-x}
+y' & = & -\dfrac{x+y+\frac{1}{2}}{x-y}
 \end{array}</math>
-At the point &thinsp; <math>\left(\frac{\pi}{4},\frac{\pi}{4}\right)</math>, we have <math>x=\frac{\pi}{4}</math> and <math>y=\frac{\pi}{4}</math>. Plugging these into our equation for <math>y'</math> gives
+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{-\frac{\pi}{4}}{\left(\frac{\pi}{4}\right)^{2}\sec^{2}\left(\frac{\pi}{4}\right)-\frac{\pi}{4}}\\
+y' & = & -\dfrac{1+0+\frac{1}{2}}{1-0} = -\frac{3}{2}.\\
-\\
 \end{array}</math>
-This means the slope of the tangent line at &thinsp; <math>\left(\frac{\pi}{4},\frac{\pi}{4}\right)</math> is <math style="vertical-align: -13px">m=\dfrac{2}{2-\pi}</math>, and a point on this line is &thinsp; <math>\left(\frac{\pi}{4},\frac{\pi}{4}\right)</math>. Using the point-slope form of a line, we have
+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-\frac{\pi}{4} & = & \frac{2}{2-\pi}\left(x-\frac{\pi}{4}\right)\\
+y-0 & = & -\frac{3}{2}\left(x-1\right)\\
 \\
-y & = & \frac{2}{2-\pi}x-\frac{\pi^{2}}{4\left(2-\pi\right)}.
+y & = & -\frac{3}{2}x+\frac{3}{2}.\\
 \end{array}</math>

@@ Line 95: / Line 95: @@
 ::<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'-\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\\