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A part of the University of Botswana's STA 102: Mathematics for Business and Social Sciences II course notes. It covers the topic of implicit differentiation, where the function y is not explicitly expressed in terms of x or vice versa. examples and formulas to find dy/dx for such functions using the chain rule and product rule.
Typology: Lecture notes
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Kesaobaka Molebatsi 2 [email protected] 242B/
March 23, 2021
(^2) Department of Statistics, University of Botswana
I (^) We have dealt with functions of the form y = x. i.e., one variable explicitly expressed in terms of the other.
I (^) Consider x^2 + y 2 = 1 or x^2 + xy + y 2 = 3 which do not give y , in terms of x or vice versa.
I (^) This is not to say that one variable cannot be solved explicitly in terms of the other.
I (^) Differentiation of these implicit functions is called implicit differentiation.
I (^) We treat y as an unknown but differentiable function of x and apply the rules of diffentiating un, uv , uv , e.t.c.
Find dydx given that x^5 + 4xy 3 − 3 y 5 = 2.
d dx
(x^5 ) + d dx
(4xy 3 ) − d dx
(3y 5 ) = d dx
⇒ 5 x^4 + 4
x
d dx (y^
(^3) ) + y 3 d dx (x)
by product rule
d dx (y^
⇒ 5 x^4 + 4
x (3y 2 ) dy ︸ ︷︷ dx︸ by chain rule
+y 3 (1)
− 3 (5)y 4 dy ︸ ︷︷ dx︸ by chain rule
⇒ 5 x^4 + 12xy 2 dy dx
⇒ 5 x^4 + 4y 3 = 15y 4 dy dx
− 12 xy 2 dy ︸ ︷︷ dx︸ I have collected like terms
Factoring dydx out and dividing by 15y 4 − 12 xy 2 both sides gives,
⇒ 5 x^4 + 4y 3 =
dy dx (15y^
(^4) − 12 xy 2 ) (7)
5 x^4 + 4y 3 15 y 4 − 12 xy 2 =^
dy dx
( (^15) y 4 − 12 xy 2 15 y 4 − 12 xy 2
1
⇒ dy dx
= 5 x
(^4) + 4y 3 15 y 4 − 12 xy 2
d dx [(x^ +^ y^ )
(^5) ] = d dx (y^
(^2) ex (^) ) (15)
⇒ 5(x + y )^4 d dx
(x + y ) ︸ ︷︷ ︸ by chain rule
= y 2 d dx
ex^ + ex^ d dx
y 2 ︸ ︷︷ ︸ by product rule
⇒ 5(x + y )^4
[ (^) d
︸ ︷︷ ︸dx^ (x) 1
= y 2 ex^ + ex^ (2y ) dydx (17)
⇒ 5(x + y )^4 + 5(x + y )^4
dy dx =^ y^
(^2) ex (^) + ex (^) (2y ) dy dx (18) ⇒ 5(x + y )^4 dy dx
− ex^ (2y ) dy dx
= y 2 ex^ − 5(x + y )^4 (19)
Factoring dydx out gives
dy dx
5(x + y )^4 − 2 yex^
= y 2 ex^ − 5(x + y )^4 (20)
⇒ dy dx
= y^
(^2) ex (^) − 5(x + y ) 4 5(x + y )^4 − 2 yex^
⇒ 2 x^5 + 2yx^4 dy dx
= x^2 (2y ) dy dx
− y 2 (2x) (26)
⇒ 2 yx^4 dy dx
− 2 yx^2 dy dx
= − 2 x^5 − 2 xy 2 (27)
⇒ dy dx
2 yx^4 − 2 yx^2
= − 2 x^5 − 2 xy 2 (28)
dy dx =^
− 2 x^5 − 2 xy 2 2 yx^4 − 2 yx^2 (29)
d dx (x
(^2) y ) + d dx (xy^
(^2) ) = d dx (6)^ (30) ⇒ x^2 dy dx
(x^2 ) ︸ ︷︷ ︸ by product rule
(y 2 ) + (y 2 ) d dx
(x) ︸ ︷︷ ︸ by product rule
⇒ x^2 dy dx
+y 2 = 0 (32)
⇒ x^2
dy dx + 2xy
dy dx =^ −^2 xy^ −^ y^
⇒ dy dx
(x^2 + 2xy ) = − 2 xy − y 2 (34)
⇒ dy dx
= −^2 xy^ −^ y^
2 x^2 + 2xy
