l2e1
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Lecture 2: Exercise 1 Explanation This is an exercise in applying the process of differentiation to several functions. It is a straightforward calculation. Hint Apply the definition of differentiation step-by-step. Answer f Ht L = t 4 + 3 t 3 - 12 t 2 + t - 6. We first define f Ht +D tL = Ht +D tL 4 + 3 Ht +D tL 3 - 12 Ht +D tL 2 + Ht +D tL - 6 f Ht +D tL =D t 4 + t 4 + 4 D tt 3 + 6 D t 2 t 2 + 4 D t 3 t + 3 D t 3 + 3 t 3 +9 D tt 2 + 9 D t 2 t - 12 D t 2 - 12 t 2 - 24 D tt + t +D t - 6 then we write,
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Ejercicio del curso de mecánica de Susskind
Transcript of l2e1
Lecture 2: Exercise 1
Explanation
This is an exercise in applying the process of differentiation to several functions. It is a straightforward calculation.
Hint
Apply the definition of differentiation step-by-step.
Answer
� f HtL = t4
+ 3 t3
- 12 t2
+ t - 6.
We first define
f Ht + D tL = Ht + D tL4+ 3 Ht + D tL3
- 12 Ht + D tL2+ Ht + D tL - 6
f Ht + D tL = D t4
+ t4
+ 4 D t t3
+ 6 D t2
t2
+ 4 D t3
t + 3 D t3
+ 3 t3
+9 D t t2
+ 9 D t2
t - 12 D t2
- 12 t2
- 24 D t t + t + D t - 6
then we write,
f Ht + D tL - f HtL = D t4
+ t4
+ 4 D t t3
+ 6 D t2
t2
+ 4 D t3
t + 3 D t3
+ 3 t3
+9 D t t2
+ 9 D t2
t - 12 D t2
- 12 t2
- 24 D t t + t + D t
-6 - t4
- 3 t3
+ 12 t2
- t + 6
f Ht + D tL - f HtL = D t4
+ 4 D t t3
+ 6 D t2
t2
+ 4 D t3
t + 3 D t3
+9 D t t2
+ 9 D t2
t - 12 D t2
- 24 D t t + D t
then we divide by D t,
f Ht + D tL - f HtL
D t
= D t3
+ 4 t3
+ 6 D t t2
+ 4 D t2
t + 3 D t2
+9 t2
+ 9 D t t - 12 D t - 24 t + 1
then we take the limit as D t ® 0,
limD t®0
f Ht + D tL - f HtL
D t
= 4 t3
+ 9 t2
- 24 t + 1.
this is the derivative.
� gHxL = sin x - cos x.
Here we using the sum rule and the results of Eq. (2),
g ' HxL = cos x + sin x.
� ΘHΑL = eΑ
+ Α ln Α.
We begin by using the sum rule
Θ ' HΑL =d
d Αe
Α+
d
d ΑHΑ ln ΑL.
We then apply Eq. (2) for the first term,
Θ ' HΑL = eΑ
+d
d ΑHΑ ln ΑL.
We then apply the product rule for the second term,
Θ ' HΑL = eΑ
+ Αd
d Αln Α + ln Α
d
d ΑΑ .
We then apply Eq. 2 for the second term,
Θ ' HΑL = eΑ
+Α
Α+ ln Α
d
d ΑΑ = e
Α+ 1 + ln Α
d
d ΑΑ
Since d Α �d Α = 1,
Θ ' HΑL = eΑ
+ 1 + ln Α.
� xHtL = sin2
t - cos t.
Here we apply the sum rule,
x ' HtL =d
d t
sin2
t -d
d t
cos t.
We complete the second term, as it is simpler,
x ' HtL =d
d t
sin2
t - H-sin tL =d
d t
sin2
t + sin t.
The second term is a little tricky. The easy way is to apply the chain rule. We define a new variable, u = sin t, then
d
d t
sin2
t =d
d u
u2
d
d t
Hsin tL = 2 u cos t
we then substitute the value of u,
2 l2e1.cdf
d
d t
sin2
t = 2 sin t cos t.
So,
x ' HtL = 2 sin t cos t + sin t.
l2e1.cdf 3
