322 Basic Engineering Mathematics
dy
dx= 3 (− 2 x−^3 )− 2 (4cos4x)+ 2 (−e−x)+1
x
=−6
x^3−8cos4x−2
ex+1
xNow try the following Practice ExercisePracticeExercise 136 Standard derivatives
(answers on page 354)- Find the gradient of the curve
y= 2 x^4 + 3 x^3 −x+4 at the points
(a)( 0 , 4 ) (b)( 1 , 8 ) - Differentiate with respect tox:
y=
2
x^2+2ln2x− 2 (cos5x+3sin2x)−2
e^3 x34.9 Successive differentiation
When a functiony=f(x)is differentiatedwith respect
tox,thedifferentialcoefficientiswrittenasdy
dxorf′(x).
If the expression is differentiated again, the second dif-ferential coefficient is obtained and is written asd^2 y
dx^2
(pronounced dee twoyby deex squared) or f′′(x)
(pronouncedfdouble-dashx). By successive differen-tiation further higher derivatives such asd^3 y
dx^3andd^4 y
dx^4
may be obtained. Thus,ify= 5 x^4 ,dy
dx= 20 x^3 ,d^2 y
dx^2= 60 x^2 ,d^3 y
dx^3= 120 x,d^4 y
dx^4=120 andd^5 y
dx^5= 0Problem 24. Iff(x)= 4 x^5 − 2 x^3 +x−3, find
f′′(x)f(x)= 4 x^5 − 2 x^3 +x− 3f′(x)= 20 x^4 − 6 x^2 + 1f′′(x)= 80 x^3 − 12 x or 4 x( 20 x^2 − 3 )Problem 25. Giveny=2
3x^3 −4
x^2+1
2 x−√
x,determined^2 y
dx^2y=2
3x^3 −4
x^2+1
2 x−√
x=2
3x^3 − 4 x−^2 +1
2x−^1 −x1
2dy
dx=(
2
3)
(
3 x^2)
− 4(
− 2 x−^3)+(
1
2)
(
− 1 x−^2)
−1
2x−1
2i.e. dy
dx= 2 x^2 + 8 x−^3 −1
2x−^2 −1
2x−1
2d^2 y
dx^2= 4 x+( 8 )(− 3 x−^4 )−(
1
2)
(
− 2 x−^3)−(
1
2)(
−1
2x−3
2)= 4 x− 24 x−^4 + 1 x−^3 +1
4x−3
2i.e.
d^2 y
dx^2= 4 x−24
x^4+1
x^3+1
4√
x^3Now try the following Practice ExercisePracticeExercise 137 Successive
differentiation (answers on page 354)- If y= 3 x^4 + 2 x^3 − 3 x+2, find (a)
d^2 y
dx^2
(b)d^3 y
dx^3- Ify= 4 x^2 +
1
xfindd^2 y
dx^2- (a) Given f(t)=
2
5t^2 −1
t^3+3
t−√
t+1,
determinef′′(t).
(b) Evaluatef′′(t)in part (a) whent=1.- Ify=3sin2t+cost,find
d^2 y
dx^2- Iff(θ )=2ln4θ, show thatf′′(θ )=−
2
θ^2