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Material Type: Assignment; Class: DIFFEREN AND INTEGRAL CALCULUS; Subject: Mathematics; University: University of Texas - Austin; Term: Fall 2008;
Typology: Assignments
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Implicit differentiation allows you to differentiate functions that are not given in the simple form y = f (x). All you have to do is treat y as a function of x and use the chain rule. For example to differentiate y^2 , we use the power rule to get 2y · dydx = 2yy′. The key to understand is that you will not be solving for y, but rather y′.
(3.5.45) Differentiate y = cos
sin(tan πx). Solution: This is a challenge problem involving the chain rule:
y′^ = − sin
sin(tan πx)
sin(tan πx)
= − sin
sin(tan πx)
(sin(tan πx))^1 /^2
= − sin
sin(tan πx)
(sin(tan πx))−^1 /^2
· (sin(tan πx))′
= − sin
sin(tan πx)
(sin(tan πx))−^1 /^2
· cos(tan πx) · (tan πx)′
= − sin
sin(tan πx)
(sin(tan πx))−^1 /^2
· cos(tan πx) · sec^2 (πx) · (πx)′
= − sin
sin(tan πx)
(sin(tan πx))−^1 /^2
· cos(tan πx) · sec^2 (πx) · π
−π sin
sin(tan πx)
· cos(tan πx) · sec^2 (πx) 2
sin(tan πx)
Solution: Differentiating both sides with respect to x, we get the chain of implications
sin(x^2 ) · (y)′^ + y · cos(x^2 )(x^2 )′^ = sin(y^2 ) · 1 + x · cos(y^2 )(y^2 )′ y′^ sin(x^2 ) + 2xy cos(x^2 ) = sin(y^2 ) + 2xyy′^ cos(y^2 ) y′(sin(x^2 ) − 2 xy cos(y^2 )) = sin(y^2 ) − 2 xy cos(x^2 )
y′^ =
sin(y^2 ) − 2 xy cos(x^2 ) sin(x^2 ) − 2 xy cos(y^2 )
Solution: We have a point, so all we need is the slope. We get this from the derivative:
(x^2 + 2xy − y^2 + x)′^ = (2)′ 2 x + (2xy)′^ − 2 yy′^ + 1 = 0 2 x + 2y + 2xy′^ − 2 yy′^ + 1 = 0 y′(2x − 2 y) = − 2 x − 2 y − 1 y′^ =
− 2 x − 2 y − 1 2 x − 2 y
Evaluating this at (x 0 , y 0 ) = (1, 2), we get that the slope at out point is y′(1, 2) = −^22 −−^44 − 1 = − −^72 = 72. Using the point-slope equation of a line, y −y 0 = y′(1, 2)·(x−x 0 ), we get y −2 = 72 (x−1). Rewriting this, we get y = 72 x − 32.
