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Unformatted text preview: Math 221 ws 1 Tutorial Program Implicit Differentiation Goal: To be able to ﬁnd derivatives using implicit diﬁerentiation
and to know when to use it. When y is a function of x implicitly: Often an equation re lates the variables in a form such as y4 + ya: + :64 = 0 rather
than the usual y : f form. In these cases will you will use
implicit differentiation. Independent and dependent variables: the form y=f(x) tells you
that “y is a function of x”<:> that X is the independent variable and
y is the dependent variable. In derivatives the instructions tell you
which is independent and which dependent: dy/dm <:> “the derivative of y with respect to x77
x is independent, y is dependent, y:f(x) dG/dt <:> t is independent, (9 is dependent, (9 : f(t)
dm/dy <=> y is independent, X is dependent, a: = g(y) How to use implicit differentiation: Step 1: Differentiate each term on both sides of the equation with
respect to the independent variable. Use all the appropriate dif—
ferentiation rules, in particular, the chain rule with the dependent
variable. Step 2: Solve the new equation for the desired derivative. The chain rule is needed with every dependent variable. d4, Example 1: Find dy/dx when y2 + my 2 1. Solution: Notice that y is the dependent variable. 2343—: W 353—: + 1 y = 224% W Had—y = *3}
dm d3: gaykw) : —y d_y : —y dm 2y+$ d3: 2 Example 2: Find a when a: — sina: = tant Solution: Notice that now X is dependent. 2 dm dx 2t
33— — cosx— 2 sec .
dt dt d
d—f(2x — cos x) 2 sec2 t dm sec2 t E _ 2x—c0sx Problems: Find dy/dx of the following: 1. 312:362—30 2. y:xvm2+1 1
3. y2=x2+$—2 Find the equation of the line tangent to each of the following curves
at the point P0. 6. $2+$yiy2=lat P0=(2,3)
7. x2y2 :9 at P0 : (—1,3) 8. The relation 372 + y2 = 25 gives the equation of a circle. Find the
slope of the tangent to the Circle at the point (0,5). 9. Find the point on the parabola (y — 4)2 : x + 2 where the
tangent is parallel to the line 305 + 6y 2 2. 10. Given the relation m2 + my + y2 = 7
a) Find the two points where is crosses the x—axis and show
that the tangents at these two points are parallel. What is
their common slope?
b) Find the points where the tangents to this curve are parallel to theX—axis.
Answers:
1 dy_2a:—l
'dxi 2y
2 22 1
2 Q2 x2+1+ x — x +
d1? m2+1 m2+l
ad—yzlAi1
d3: 33%
4 dy_y—ac(ac+y)2_1—3m2—2a:y
dl’— a: _ 1+$2
5 d$_$(liy2)
'dy—y(w21)
6 7m—4y:
7. 3m—y:—6 8. slope is 0 9.(—1,3) 10. (ﬁﬁ) and _ 0 =
7 a d‘r (\/:7 2 ;)and (W > d—y —2
(m; ...
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 Spring '06
 OnlineResources
 Implicit Differentiation

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