MAC 2311C Lecture 10: 3.5 :implicit Differentiation
22 Jun 2018
School
Department
Course
Professor
3. 5
1
Y
COS (×)=4×2 +2y2
¥(ycos (×))=£(4×2 +zyz )
If product
-
ysin + cos 'Adat× =8x +4g (IF )
cos (x)£¥ -tgddt )=sx
+gsin
¥f (cos (x)-4g )=8× +
ysin (x)
Tt =8×tys=)
COS (×)-4g
2
y
= arccos (e2× )
y
' =¥cos
-'
(ezx )£e2× £
2×
=He
(e2× )(2) =
¥eIx
3EYCOSCX )=5+sin (xy )
€× EYCOS (×)=£(S+sincxy ))
eYcoscx5tCoscx7e41-adzsincxy7adxCxy7-eYsinCx7tcoscx7eYddIt-cosCxy7CxyIyx11-eYsincxstcosCx7eYddIt-CosCxy7CxfItyl-eYsincxItCoscx1eYadF-XTYcoscxy7tycosCxy1coscx7eYddTT-XddIrCosCxy-eYsincx7tyCosCxyddTCcosCx1eY-xcoscxy7-eYsincx7tycoscxDfga-e7sinC-CsyCoSCx7eY-XcosCxyj@fCx1t.X
2[ f(×)]
"
=\ 8f( i)=2
-
Product rule &chain rule
f'(1)+×2 (fci
))4'
+€
((7)
"
×
21 =Ix 2
f/( 1) +×
2
(4)(f( ,))3 (f/( 1))+f(i)
4
ZX =0
+4(g)f'(1))+16 (2) =0
32 f '(1)+32 = 0
f'CI)t32 f'(1)=-32
33 f '(I)=-32
fcc ()=-
3233
52×3 +×2y -xy
3
=3
,
,
£2×3 +(×2j
+ y
'x2 )-(xy3 + xy 3)
6×2 +(Zxy +IF ×
')-(x 3y2a±f + y3) =0
6×2 + Zxy + ×2 }×d -×3y2 Tf y
3
=0
6×2 + zxy y3 =-x2¥t×+ ×3Y2 IF
day =6×2+2×-83
6×2+2 xy y3
=IF (-×2 + ×3y2 )-xz + xsyz
6e×" =zx -
y
£,
e×' 4=£(2× -g) year .×e×' 8If = zytadtfyz
adz±y (ex
'Y)= 2-¥8 ye
"Y -2y2
=-F×tyz + xexar adds
giggly
'
(end )= 2-IF Ye "Y
-2y2
=If (×e×4 -yr )
8- ×a¥ (exiajitz
.day (yz )Ttx
=YIItyIgE@
×8 + y3=5 3y2
lgyz
=8×
¥× (x8+y3 )=Fx 5day
=
-
8×-7
8×7
+ 342 #3g2
dx
8sxty =9+x2y2
(E) (sxtyt :( s
'
+If )=×"y2 +g"x2
=2xy2 +2g IT x2
25K¥54 =rxyz +zyatex '
s+Ff =uxytsxtf +ty Ff ×Tsftiy
IF -4gE#xtj= ↳ xyzrsxty -s
a±f( I-xistxy +4g
)=uxyifxy -S
T# =4xyft.gs
l-txiysxtty
