<Exercise 3> 풀이

Q1. (a) The function [eqn: y=17] is a horizontal line, indicating that y equals 17 no matter what value x assumes. Hence a small change in x, or any change in x, produces no change in y. If [eqn: dy=0] for any change in x, then [eqn: dy/dx] (the derivative) must also equal zero. constant rule.

(b) [eqn: f(x)=5x+12].  [eqn: y prime =5].  linear function rule.

(c) [eqn: y=5x ^{2}]. [eqn: d over dx (5x sup 2 ) = 10x].

At [eqn: x=2], the slope [eqn: dy/dx=10(2)=20],

At [eqn: x=5], [eqn: dy/dx=50]. power function rule.

(d) [eqn: f(x)=5x ^{3} -7x ^{2} +4x+9]. [eqn: f _{x} =15x ^{2} -14x+4].

sums and differences rule.

Q2. given [eqn: y=5x ^{4} (3x-7)]

(a) product rule. [eqn: y=uv.`````````````dy/dx=uv prime +vu prime]. Let [eqn: u=5x ^{4}] and [eqn: v=3x-7]. Then [eqn: u prime =du/dx=20x ^{3} ,]. and [eqn: dv/dx=3]. Substitute these values in the product rule, [eqn: {dy} over {dx} `=5x ^{4} (3)+(3x-7)(20x ^{3} )].

Simplifying algebraically,

[eqn: {dy} over {dx} =15x ^{4} +60x ^{4} -140x ^{3} =75x ^{4} -140x ^{3}].

(b) Simplifying the original function through

multiplication, [eqn: y=5x ^{4} (3x-7)=15x sup 5 -35x sup 4].

Taking the derivative, [eqn: y prime =75x ^{4} -140x ^{3}].

Q3. quotient rule.

(a) [eqn: y= {3x ^{8} -4x ^{7}} over {4x ^{3}}]. let [eqn: u=3x ^{8} -4x ^{7}](numerator),

[eqn: v=4x ^{3}](denominator). Thus, [eqn: du/dx=24x ^{7} -28x ^{6}],

[eqn: dv/dx=12x ^{2}]. Substituting in the quotient rule,

[eqn: {dy} over {dx} = {4x ^{3} (24x ^{7} -28x ^{6} )-(3x ^{8} -4x ^{7} )(12x ^{2} )} over {(4x ^{3} ) ^{2}}# `````````````= {96x ^{10} -112x ^{9} -36x ^{10} +48x ^{9}} over {16x ^{16}} = {60x ^{10} -64x ^{9}} over {16x ^{6}}# ````````````= {15} over {4} x ^{4} -4x ^{3}].

(b) [eqn: y= {4x ^{5}} over {1-3x}]  (note: [eqn: x != 1/3],  if [eqn: x=1/3], the

denominator would equal zero and the function would be undefined).

[eqn: {dy} over {dx} = {(1-3x)(20x ^{4} )-4x ^{5} (-3)} over {(1-3x) ^{2}} = {20x ^{4} -48x ^{5}} over {(1-3x) ^{2}}].

(c) [eqn: y= {5x ^{2} -9x+8} over {x ^{2} +1}].

[eqn: {dy} over {dx} = {(x ^{2} +1)(10x-9)-(5x ^{2} -9x+8)(2x)} over {(x ^{2} +1) ^{2}}# ````````= {10x ^{3} -9x ^{2} +10x-9-10x ^{3} +18x ^{2} -16x} over {(x ^{2} +1) ^{2}} = {9x ^{2} -6x-9} over {(x ^{2} +1) ^{2}}]

Q4. chain rule.

(a) [eqn: y=u ^{20} ,````````u=7x-4], [eqn: {dy} over {dx} =20u ^{19} (7)=140u ^{19}].

But since [eqn: u=7x-4], [eqn: {dy} over {dx} =140(7x-4) ^{19}].

(b) [eqn: y=(x ^{2} +3x-1) ^{5}].  Let [eqn: u=x ^{2} +3x-1].

Then [eqn: y=u ^{5} ,````````dy/du=5u ^{4}], and  [eqn: du/dx=2x+3].

Substituting in the chain rule formula,

[eqn: dy/dx=5u ^{4} (2x+3)=(10x+15)u ^{4}# ```````````````````````=(10x+15)(x ^{2} +3x-1) ^{4}].

Q5. product rule + quotient rule.

(a) [eqn: y= {3x(2x-1)} over {(3x-2)}].

Let [eqn: u=3x(2x-1),``````v=3x-2],  [eqn: {dy} over {dx} = {vu prime -uv prime } over {v ^{2}}],

and [eqn: dv/dx=3]. From the product rule,

[eqn: du/dx=3x(2)+(2x-1)(3)=12x-3].

Thus  [eqn: {dy} over {dx} = {(3x-2)(12x-3)-3x(2x-1)(3)} over {(3x-2) ^{2}}].

Simplifying algebraically,

[eqn: {dy} over {dx} = {36x ^{2} -9x-24x+6-18x ^{2} +9x} over {(3x-2) ^{2}} = {18x ^{2} -24x+6} over {(3x-2) ^{2}}].

(b) [eqn: y=3x[ {(5x+2)} over {(2x+3)} ]].

Let [eqn: u=3x,``````v=(5x+2)/(2x+3)],

then [eqn: {dy} / {dx} =uv prime +vu prime] and [eqn: u' = du/dx=3].

From the quotient rule,

[eqn: {dv} over {dx} = {(2x+3)(5)-(5x+2)(2)} over {(2x+3) ^{2}} = {11} over {(2x+3) ^{2}}].

Substituting in the product rule formula,

[eqn: {dy} over {dx} =3x {11} over {(2x+3) ^{2}} + {5x+2} over {2x+3} (3)# `````````````= {33x+(15x+6)(2x+3)} over {(2x+3) ^{2}} = {30x ^{2} +90x+18} over {(2x+3) ^{2}}]

(c) 과정생략. [eqn: y prime ={630x ^{3} +135x ^{2} +30} over {25x ^{2}}].

Q6. second-order derivative, notations.

(a) [eqn: y=7x ^{3} +5x ^{2} +12].  [eqn: y prime =21x ^{2} +10x],

[eqn: y '' = {d ^{2} y} over {dx ^{2}} =42x+10], at [eqn: x=2], [eqn: y '' rm `=42(2)+10=94].

(b) [eqn: y= {5x} over {1-3x}]. [eqn: D sub y`= {(1-3x)(5)-5x(-3)} over {(1-3x) ^{2}} = {5} over {(1-3x) ^{2}}]

[eqn: D sub y sup 2 = {(1-3x) ^{2} (0)-5[2(1-3x)(-3)]} over {(1-3x) ^{4}} = {30-90x} over {(1-3x) ^{4}}].

at [eqn: x=2], [eqn: y '' =-6/25].

(c) [eqn: y prime =150x ^{5} -350x ^{4} +196x ^{3}],

[eqn: y '' =750x ^{4} -1400x ^{3} +588x ^{2}], at [eqn: x=2], [eqn: D _{y} ^{2} =3152].

Q7. successive derivatives.

(a) [eqn: y=x ^{3} +3x ^{2} +9x-7].

[eqn: y prime =3x ^{2} +6x+9`],	at [eqn: x=3], [eqn: y prime =3(3) ^{2} +6(3)+9=54]

[eqn: y prime prime =6x+6],	at [eqn: x=3], [eqn: y prime prime =6(3)+6=24]

[eqn: y prime prime prime =6],	at [eqn: x=3], [eqn: y prime prime prime =6]

[eqn: y ^{(4)} =0],	at [eqn: x=3], [eqn: y ^{(4)} =0]

(b)[eqn: y=(4x-7)(9x+2)]

[eqn: y prime =72x-55],	at [eqn: x=3], [eqn: y prime =161]

[eqn: y prime prime =72],	at [eqn: x=3], [eqn: y prime prime =72]

[eqn: y prime prime prime =0],	at [eqn: x=3], [eqn: y prime prime prime =0]

(c) [eqn: y=(5-x) ^{4}]

[eqn: D _{y} =-4(5-x) ^{3} ,````````````````````at``````x=3,``````D _{y} =-32# D _{y} ^{2} =12(5-x) ^{2} ,``````````````````````at``````x=3,``````D _{y} ^{2} =48# D _{y} ^{3} =24x-120,`````````````````````````at``````x=3,``````D _{y} ^{3} =-48# D _{y} ^{4} =24,`````````````````````````````````````````````````at``````x=3,``````D _{y} ^{4} =24# D _{y} ^{5} =0,`````````````````````````````````````````````````````at``````x=3,``````D _{y} ^{5} =0]