Higher-Order Derivatives and Implicit Differentiation solved questions

CHAPTER 12
Higher-Order Derivatives
and Implicit Differentiation


12.1 Find the second derivative y" of the function by direct computation.

By the chain rule, By the


quotient rule,


12.2 Use implicit differentiation to solve Problem 12.1.

y2 = x2 + 1. Take the derivative of both sides with respect to x. By the chain rule,
Thus, 2yy' = 2x, and, therefore, yy' =x. Take the derivative with respect to x of both sides, using the
product rule on the left: yy" + y'-y' — \. So, yy" = 1 - (y')2. But, since yy'— x, y' = xly. Hence,
yy" = 1 - x2/y2 = (y 2 - x 2 ) / y 2 = 1 ly2 = 1 /(x2 + 1). Thus, y" = 1 ly(x2 + 1) = 1 l(x2 + I)3'2.

12.3 Find all derivatives y'"' of the function y = irx3 — Ix.
y' = 3irx2-7, y" = 6trx, y'" = 6ir, and y ( n ) = 0 for n>4.

12.4 Find all derivatives y(n) of the function
and
This is enough to detect the general pattern:

12.5 Find all derivatives y <n) of the function y = 1 /(3 + x).
y = (3 + *r'




The general pattern is


12.6 Find all derivatives y ( "'of the function y = (x + l ) / ( x - 1).




The general pattern is


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12.7 Find all derivatives yw of the function
Use implicit differentiation, y =2x-l. Hence, 2yy' = 2, y ' = y '. So,

y" = -y~2 • y' = -y~2 -y~1 = -y~3
y>» = 3y-*.y>=3y-<.y->=3y-S


yw = -3 • 5y~6 -y' = -3- 5y"6 • y~l = -3 • 5y~7
So, the pattern that emerges is



12.8 Find all derivatives yM of the function y = sin x.
y' = cos x, y" = — sin x, y'" = -cos x, y <4> = sin x, and then the pattern of these four functions keeps on
repeating.

12.9 Find the smallest positive integer n such that D"(cos x) - cos x.
Let y = cosx. y'= — sinx, y"=—cosx, y'" = smx and 3* —cos*. Hence, n=4.

12.10 Calculate >> <5) for y = sin2 x.
By the chain rule, y' = 2 sin x cos x = sin 2*. Hence, y" = cos 2x • 2 = 2 cos 2x, y'" = 2(-sin 2x) • 2 =
-4 sin 2x, yw = -4 (cos 2x) • 2 = -8 cos 2x, >-C5) = -8(-sin 2x) • 2 = 16 sin 2x = 16(2 sin x cos x) = 32 sin x cos x.

12.11 On the circle x2 + y2 = a2, find y".

By implicit differentiation, 2x + 2yy'=0, y'=—x/y. By the quotient rule,



12.12 If x 3 -/ = l, find/'.
Use implicit differentiation. 3x2-3y2y' = Q. So, ;y' = ;t2/}'2. By the quotient rule,




12.13 If xy + y2 = l, find y' and y".
Use implicit differentiation, xy' + >> + 2yy' =0. Hence, yX* + 2y) = -y, and y' = By
the quotient rule,




12.14 At the point (1,2) of the curve x2 — xy + y2 =3, find an equation of the tangent line.
I Use implicit differentiation. 2x - (xy' + y) + 2yy' = 0. Substitute 1 for x and 2 for y. 2 - (y' + 2) + 4y' =
0. So, y ' = 0 . Hence, the tangent line has slope 0, and, since it passes through (1,2), its equation is y = 2.

12.15 If x2 + 2xy + 3y2 = 2, find y' and /' when y = l.