# 有幾題數學唔識 urgent!!

1.) Find dy/dx, using implicit differentiation

6xy-(y/6)=(4/x)

dy/dx=

2.)Use implicit differentiation to find the following.

2x^2-y^2=xy (-1,2)

a.)The slope of the tangent line at the indicated point on the graph

m=

b.)The equation of the tangent line at the indicated point on the graph.

y=

3.) An employment research company estimates that the value of a recent MBA graduate to an accounting company is given by the equation below, where V is the value of the graduate, e is the number of years of prior business experience, and g is the graduate school grade-point average. If V is fixed at 250 , find de/dg when g=3.1 (Round the answer to 2 decimal places.)

V=3e^2+3g^3

a.) de/dg=

b.) This means that, at a value level of 230, and increase by 1.0 for a candidate with a 3.1 grade point average is worth ( ) years of experience.

呢幾題搞到我頭都大晒

請好心人幫我解答

感激! ^^

### 1 Answer

- 1 decade agoFavorite Answer
1.) 將兩邊differentiate with respect to x

d[6xy-(y/6)]/dx = d[4/x]/dx

6y-(1/6)dy/dx = -4(1/x^2)

dy/dx = 36y+24/x^2

2a.) slope of tangent at (-1,2) = dy/dx at (-1,2) = m

所以搵到dy/dx再sub番(-1,2)即可

2x^2-y^2 = xy

一樣先將兩邊diff w.r.t. x

4x-d[y^2]/dx = y

4x-(d[y^2]/dy)(dy/dx) = y (chain rule)

4x-2y(dy/dx) = y

dy/dx = (4x-y)/2y

m = (4(-1)-2)/2(2) = -3/2

2b.) 果條 tangent line (straight line)既slope係m, pass through (-1,2)

於是我地可以用中學教過既(XD) point-slope form

which is y-y1 = m(x-x1)

=> y-2 = (-3/2)(x+1)

y = (-3x+1)/2

3a.) 都係將兩邊d wrt g先, 而因為搵change of e wrt g唔關V事, 所以treat V as constant

V=3e^2+3g^3

=> 0 = (d[3e^2]/de)(de/dg)+9g^2

0 = 6e(de/dg)+9g^2

de/dg = (-9g^2)/6e

3b.) 要搵de/dg at V=230, g=3.1

(你上面寫V=250, 下面寫V=230, 我當係230先la, 如果唔係你跟step計番一次就ok)

但係我地上面計到既de/dg係in terms of g同e

所以要計左e出黎先 (用原本果條式)

V=3e^2+3g^3

230=3e^2+3(3.1)^3

e=6.8465...

代入de/dg入面

=> de/dg = (-9g^2)/6e

de/dg = (-9(3.1)^2)/6(6.85) = -2.11

所以

This means that, at a value level of 230, and increase by 1.0 for a candidate with a 3.1 grade point average is (roughly...) worth ( 2.11) years of experience.

因為當V fix死左係230, at g=3.1, 個rate of change of e wrt g係-2.11, 所以approximately when g changed by 1, e 會changed by -2.11. 亦即係話當你gpa係3.1, increase 1 gpa 可以令你少2.11 work experience 就attain到230既value level.

(for your reference: 當V fix係一個value時原本果條equation就係一條isoquant, 所以de/dg 就係所謂既 marginal rate of substitution)

step應該冇錯, 但可能會計錯, 你最好計多次 (因為我成有呢d大意野XD)

2009-02-22 06:29:05 補充：

sorry for 中英夾雜...我諗咁樣會易明d

2009-02-22 13:42:35 補充：

啊你的account是奇摩的

看得懂廣東話嗎?

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