\magnification 2000
\parskip 10pt
\parindent 0 pt
\def\c{\vskip 1.5cm}
\def\b{\vfill}
\def\u{\vskip -10pt}
\def\v{\vskip 10pt}
\vsize 9.5 truein
%%\voffset 0.3truein
\hoffset -0.5truein
\hsize 6.6 truein


By the chain rule $[(x^2+1)^{9}]' = 9(x^2+1)^{8}(2x)$

Or in other words,  ${d [(x^2+1)^{9}] \over  dx} = 9(x^2+1)^{8}(2x)$
 
 Or in other words,  if we let $u = x^2+1$, then 
 \v
 \centerline{${du \over dx} = u' = 2x$ and}
 \v
\centerline{$[(x^2+1)^{9}]'  = [u^{9}]' = 9u^{8}u' = 9(x^2+1)^{8}(2x)$}
 
 Or in other notation, 
 \v
\centerline{${d [(x^2+1)^{9}] \over  dx} =  {d [u^{9}] \over  dx} = 9u^{8}{du \over dx} = 
9(x^2+1)^{8}(2x)$}
 

We can use the chain rule to calculate a derivative using implicit differentiation.



Ex:  Find the slope of the tangent line to $x^2 + y^2 = 1$ at the point $(x, y) = ({1 \over \sqrt{2}}, -{1 \over \sqrt{2}})$.


Long method (without using implicit differentiation):

Solve for $y$:

 $x^2 + y^2 = 1$ 
 implies
  $y^2 = 1 - x^2$ 
 implies
 $y = \pm \sqrt{1 - x^2}$ 


Since the $y-$value of the point  $(x, y) = ({1 \over \sqrt{2}}, -{1 \over \sqrt{2}})$ is negative, we are interested in the bottom half of the circle:


\centerline{$y =  - \sqrt{1 - x^2} =  -(1 - x^2)^{1 \over 2} $}

To find slope of the tangent line, take derivative:  


\v
\centerline{${dy \over dx} = -{1 \over 2} (1 - x^2)^{-1 \over 2}(-2x)
= {x \over \sqrt{1-x^2}}$ }


Hence when $x = {1 \over \sqrt{2}}$, then the slope of the tangent line is

 %%\v
\centerline{${ {1 \over \sqrt{2}} \over \sqrt{1-( {1 \over \sqrt{2}})^2}}$
= ${ {1 \over \sqrt{2}} \over \sqrt{1-( {1 \over {2}})}}$
= ${ {1 \over \sqrt{2}} \over \sqrt{ {1 \over {2}}}}$
= ${ \sqrt{{1 \over 2}} \over \sqrt{ {1 \over {2}}}} = 1$}


%%Thus the equation of the tangent line at the point $({1 \over \sqrt{2}}, -{1 \over \sqrt{2}})$ 
%%is ${y + {1 \over \sqrt{2}}) \over x - {1 \over \sqrt{2}} = 1 $ 


\v
\hrule

We can instead use implicit differentiation:

Note that $y$ is a function of $x$ for the bottom half of the circle:  $y = f(x) = -(1 - x^2)^{1 \over 2} $


Thus to find the derivative of $y^2$, we can use the chain rule:

\v\centerline{${d(y^2) \over dx} = 2y \cdot {dy\over dx} = 2(-(1 - x^2)^{1 \over 2})\cdot {x \over \sqrt{1-x^2}} = -2x$}


Note $y^2 = [-(1 - x^2)^{1 \over 2}]^2 = 1 - x^2$

However, we don't need  the above to find the slope of the tangent line to the unit circle at 
the point $(x, y) = ({1 \over \sqrt{2}}, -{1 \over \sqrt{2}})$.  Instead:
\vfill
\eject
{\bf Shorter method} for finding this slope of the tangent line to $x^2 + y^2 = 1$ at
the point $(x, y) = ({1 \over \sqrt{2}}, -{1 \over \sqrt{2}})$.

We have  $x^2 + y^2 = 1$, and we want to find slope = ${dy \over dx}$

Take the derivative with respect to $x$ of both sides:

$${d(x^2 + y^2) \over dx} = {d(1) \over x}$$

$$2x + 2y \cdot {dy\over dx} = 0$$

Solve for $ {dy\over dx}$:  ~~~~~~~~$2y \cdot {dy\over dx} = -2x$

$$ {dy\over dx} = {-x \over y}$$

Hence the slope  of the tangent line to the unit circle at 
the point $(x, y) = ({1 \over \sqrt{2}}, -{1 \over \sqrt{2}})$. 

$$ {dy\over dx} = {-x \over y} =  {-{1 \over \sqrt{2}} \over-{1 \over \sqrt{2}}} = 1$$


~\vfill


\eject
Suppose $2x^2y - 3y^2 = 4$.  First find $y'$:

Easiest method is to use implicit differentiation.  Take derivative (with 
respect to $x$) of both sides.
\v
\centerline{${d \over dx}(2x^2y - 3y^2) = {d \over dx}(4)$}
\v
\centerline{$4xy + 2x^2y' - 6yy' = 0$}

Solve for $y'$ (note this step is easy as one can factor $y'$ from some 
terms.  Observe that this will always be the case):
\v

\centerline{$y'(2x^2 - 6y) = -4xy$}
\v

\centerline{$y'= {-4xy \over 2x^2 - 6y} ={-2(2xy) \over -2(3y - x^2)} = 
{2xy \over 3y - x^2}$}

Hence $y'= {2xy \over 3y - x^2}$







\end


Proctoring 1:30pm exam: Grant Gallagher, Julia

Proctoring 2:30pm exam: Michael Covello, Williams, Tristan J N, Julia

Picking up exam to distribute to the 2:30pm section:
versions 1 & 4: Williams, Tristan J N
versions 2 & 5: Michael Covello
versions 3 & 6: ?

Collating finished exams at 4:30pm in 110MLH: Michael Covello, Julia

