Solutions to the exercises

 

A1) = (a – x)/x

A2) = 1 + y

A3) =

A4) = A²/CB

A)5 = cA/(A +c)U²

A6) = lna

A7) = d*lnc

A8) There are several ways of solutions. Most simply substitute e high -ax into u and solve after u first.

a) x = 1/a*ln[b/(b-1)] or x = -1/a*ln[(b – 1)/b] what is exactly the same.

b) put that solution into the equation you solved. Take attention that -(b – 1) = -b + 1!

A9) First put this equation with basis e

e high lnx = e high (-B + lnC²)

the you get

x = e high (-B + lnC²) = C²*e high -B

A10) = ln(e*a/b), (lne¹ = 1!)

A11) ax² + bx + c = 0,

D = b² – 4ac

x1 = (-b + √D)/2a

x2 = (-b – √D)/2a

A19)

D = a1*b2 – b1*a2

D1 = c1*b2 – c2*b1

D2 = a1*c2 – c1*a2

x = D1/D

y = D2/D

A20) Open excel.

Choose e.g. the Fields: A1 = a1, B1 = b1 uns C1 = c1

Excelformula for x  =(C1*B2-C2*B1)/(A1*B2-B1*A2)

Excelformula for y =(A1*C2-C1*A2)/(A1*B2-B1*A2)

A21)

x = -1

y = 2

A22) D = 0 however D1 and D2 are different from zero. This equation system has no solution.

A23) Here all of the 3 determinants, D, D1 and as well D2 are Zero, = 0. The excelformula just shows the result “#DIV/0!” (like in the exercise above)

But attention please: This equation system has infinitely many solutions!

 

Solution of 1
G‘(u) = ehigu(sinu + cosu)

Solution of 2
J‘(v) = 3v² + 10v, J‘(3) = 57

Solution of 3

Deriving the fraction using the quotient rule:

= (-2Ex + 2r² + 2E² – r²)/(E – r)²
Deriving of 2E*ln(E – r) by chain rule:

= -2E/(E-r)

Add up the two derived results by giving the denominator the same name: r²/(E – r)²

Solution of 4

1) F(x) = x³ – 2x² + 5 has 2 turning points with slope = 0. (X/Y) (0/5) and (4/3/3.81)

2) f ‘ (x) = 3x² – 4x = 0, x1 = 0, x2 = 1.3333 = 4/3

3) f (x) has one zero place: x = – 1.2419

4) ∫f(x) = x³ – 2x² + 5 = xsqr(4)/4 – 2/3x³ + 5x
[xhoch4/4 – 2/3x³ + 5x](0 – 1.2419) = |4.338 cm²|

If you manually add up the mm²-squares you should get 4338 mm² of them: 4338 mm² = 4.338 cm²

 

D)29

y = M + (N – M)/[1 – L*e hoch b(N – M)x]

M/dx = 0, thus M disappears and it remains:

dy/dx = (N – M)/[1 – L*e hoch b(N – M)x]/dx 

Chain rule:

outer function : f(u) = (N – M)/u, df(u)/du or f'(u) = -(N – M)/u²

Inner function = u = 1 – L*e hoch b(N – M)x Derivation after dx = -Lb(N – M)e high b(N – M)x

Outer multiplied with the inner function = -(N – M)*-[Lb(N – M)e high b(N – M)x]/[1 – L*e high b(N – M)x]²

So we have for the left side: y’ =  [L*b(N – M)²*e high b*(N – M)x]/[1 – L*e high b*(N – M)x]² 

The right side: b(M – y)*(B – y) we now put here y = M + (N – M)/[1 – L*e high b(N – M)x].

For convenience, we temporarily replace 1 – L*e hoch b(N – M)x with z and write: y = M + (N – M)/z.

By inserting into b(M – y)*(B – y) and multiplying out it remains -b/z(M – N)² + b/z²(M – N)² which can be simplified as follows:

= (M – N)²b(1 – z)/z² or because (M – N)² = (N – M)² = (N – M)²b(1 – z)/z²  and now we write out z.

=(N – M)²b[1 – (1 – Le high [b(n – M)x]/[1 – L*e high b(N – M)x]² = [L*b( N – M)²*e high b*(N – M)x]/[1 – L*e high b*(N – M)x]²

 

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