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  إجابات كتاب التمارين

إجابات كتاب التمارين

قاعدة السلسلة

أجد مشتقة كل اقتران مما يأتي:

(1) f(x) = 100e-0.1x

f ’(x) = -10e-0.1x

(2) f(x) = sin (x2 + 1)

f ’(x) = 2x cos (x2 + 1)

(3) f(x) = cos2 x

f ’(x) = -2 cos x sin x = - sin 2x

(4) f(x) = cos2x – 2 cos x

f ’(x) = -2 sin 2x + 2 sin x

(5) f(x) = log3 begin mathsize 20px style fraction numerator x square root of x space minus space 1 end root over denominator 2 end fraction end style

f(x) = log3 x + begin mathsize 20px style 1 half end style log3 (x – 1) – log3 2

f ’(x) = begin mathsize 20px style fraction numerator 1 over denominator x space ln space 3 end fraction end style + begin mathsize 20px style fraction numerator 1 over denominator 2 left parenthesis x space minus space 1 right parenthesis space ln space 3 end fraction end style

(6) f(x) = 2cot2 (πx + 2)

f(x) = 2(cot (πx + 2))2

f ’(x) = -4π cot (πx + 2) csc2 (πx + 2)

(7) f(x) = log 2x

f ’(x) = begin mathsize 20px style fraction numerator 2 over denominator 2 x space ln 10 end fraction end style = begin mathsize 20px style fraction numerator 1 over denominator x space ln 10 end fraction end style

(8) f(x) = ln (x3 + 2)

f ’(x) = begin mathsize 20px style fraction numerator 3 x squared over denominator x cubed space plus space 2 end fraction end style

(9) f(x) = (begin mathsize 20px style fraction numerator x squared over denominator x cubed space plus space 2 end fraction end style)2

f ’(x) = 2 x begin mathsize 20px style fraction numerator x squared over denominator x cubed space plus space 2 end fraction end style x begin mathsize 20px style fraction numerator 2 x space left parenthesis x cubed space plus space 2 right parenthesis space minus space 3 x to the power of 4 over denominator left parenthesis x cubed space plus space 2 right parenthesis squared end fraction end style

        = begin mathsize 20px style fraction numerator 2 x squared over denominator x cubed space plus space 2 end fraction end style x begin mathsize 20px style fraction numerator 4 x space minus space x to the power of 4 over denominator left parenthesis x cubed space plus space 2 right parenthesis squared end fraction end style = begin mathsize 20px style fraction numerator 8 x cubed space minus space 2 x to the power of 6 over denominator left parenthesis x cubed space plus space 2 right parenthesis cubed end fraction end style

(10) f(x) = x2 begin mathsize 20px style square root of 20 space minus space x end root end style

f ’(x) = x2 x begin mathsize 20px style fraction numerator negative 1 over denominator 2 space square root of 20 space minus space x end root end fraction end style + 2x begin mathsize 20px style square root of 20 space minus space x end root end style

         = begin mathsize 20px style fraction numerator negative x squared over denominator 2 space square root of 20 space minus space x end root end fraction end style + 2x begin mathsize 20px style square root of 20 space minus space x end root end style = begin mathsize 20px style fraction numerator 80 x space minus space 5 x squared over denominator 2 space square root of 20 space minus space x end root end fraction end style

(11) f(x) = begin mathsize 20px style fraction numerator sin space left parenthesis 2 x space plus space 1 right parenthesis over denominator e to the power of x squared end exponent end fraction end style

f ’(x) = begin mathsize 20px style fraction numerator 2 e to the power of x squared end exponent space cos space left parenthesis 2 x space plus space 1 right parenthesis space minus space 2 x e to the power of x squared end exponent space sin space left parenthesis 2 x space plus space 1 right parenthesis space space over denominator e to the power of 2 x squared end exponent end fraction end style

        = begin mathsize 20px style fraction numerator 2 space cos space left parenthesis 2 x space plus space 1 right parenthesis space minus space 2 x space sin space left parenthesis 2 x space plus space 1 right parenthesis space space over denominator e to the power of x squared end exponent end fraction end style

(12) f(x) = begin mathsize 20px style 3 to the power of cot space x end exponent end style

f ’(x) = - (begin mathsize 20px style 3 to the power of cot space x end exponent end style  ln 3) csc2 x

 

أجد معادلة المماس لكل اقتران مما يأتي عند قيمة x المعطاة:

(13) y = 2 sin 5x – 4 cos 3x , x = begin mathsize 20px style straight pi over 2 end style

begin mathsize 20px style fraction numerator d y over denominator d x end fraction end style = 10 cos 5x + 12 sin 3x

ميل المماس:

begin mathsize 20px style right enclose fraction numerator d y over denominator d x end fraction end enclose subscript space left parenthesis x equals straight pi over 2 right parenthesis end subscript end style  = -12

عندما  x = begin mathsize 20px style straight pi over 2 end style ، فإن y = 2

معادلة المماس:

  y – 2 = -12 (x - begin mathsize 20px style straight pi over 2 end style)   →  y = -12x + 6π + 2

(14) f(x) = (x2 + 2)3 , x = -1

f ’(x) = 6x (x2 + 2)2

ميل المماس:

f ’(-1) = -54

x = -1 →  y = f(-1) = 27

معادلة المماس:

  y – 27 = -54 (x + 1)  →   y = -54x - 27

(15) f(x) = tan 3x , x = begin mathsize 20px style straight pi over 4 end style

f ’(x) = 3 sec2 3x

ميل المماس:

f ’(begin mathsize 20px style straight pi over 4 end style) = 6

x = begin mathsize 20px style straight pi over 4 end style →  y = f(begin mathsize 20px style straight pi over 4 end style) = -1

معادلة المماس:

 y + 1 = 6 (x - begin mathsize 20px style straight pi over 4 end style)   →  y = 6x - begin mathsize 20px style fraction numerator 3 straight pi over denominator 2 end fraction end style - 1

 

إعداد : شبكة منهاجي التعليمية

17 / 10 / 2022

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