Find dydx for the following cases : 6(i) 2 3 4[ ( sin ) ]y x x x= + +(ii) 4 4 16x y+ =3. (a) Evaluate the following integrals : 6(i) 2 2921(2 1)t t t dtt+ −∫(ii) 3 / 20 sin x dxπ∫(b) Find dydx , when x xy x xe= + . 44. (a) Find all the roots α, β, γ of the cubicequation 3 7 6 0x x− − = . Also, find theequation whose roots are α + β, β + γand α + γ. 5

To find dy/dx for the given cases:

(i) For the equation 6x^2 + 3x^4 + 4sin(x)y = 0, we need to differentiate both sides of the equation with respect to x.

Differentiating 6x^2 with respect to x gives 12x.

Differentiating 3x^4 with respect to x gives 12x^3.

To differentiate 4sin(x)y with respect to x, we need to use the product rule. The derivative of sin(x) with respect to x is cos(x), and the derivative of y with respect to x is dy/dx. So, the derivative of 4sin(x)y with respect to x is 4cos(x)y + 4sin(x)(dy/dx).

Putting it all together, we have 12x + 12x^3 + 4cos(x)y + 4sin(x)(dy/dx) = 0.

(ii) For the equation 4x^4 + 16xy^3 = 3, we need to differentiate both sides of the equation with respect to x.

Differentiating 4x^4 with respect to x gives 16x^3.

To differentiate 16xy^3 with respect to x, we need to use the product rule. The derivative of x with respect to x is 1, and the derivative of y^3 with respect to x is 3y^2(dy/dx). So, the derivative of 16xy^3 with respect to x is 16y^3 + 48xy^2(dy/dx).

Putting it all together, we have 16x^3 + 16y^3 + 48xy^2(dy/dx) = 0.

(a) To evaluate the integral 2t^2 + 9t + 21/(2t + 1) dt, we can use the method of partial fractions.

First, we factorize the denominator: 2t + 1 = (2t + 1).

Next, we express the integrand as a sum of partial fractions:

2t^2 + 9t + 21/(2t + 1) = A/(2t + 1) + B/(2t + 1)^2.

To find the values of A and B, we can equate the numerators:

2t^2 + 9t + 21 = A(2t + 1)^2 + B(2t + 1).

Expanding and simplifying, we get:

2t^2 + 9t + 21 = 4At^2 + 4At + A + 2Bt + B.

Comparing the coefficients of like terms, we have:

4A = 2, 4A + 2B = 9, A + B = 21.

Solving these equations, we find A = 1/4 and B = 19/4.

Now, we can integrate the partial fractions:

∫(2t^2 + 9t + 21)/(2t + 1) dt = ∫(1/4)/(2t + 1) dt + ∫(19/4)/(2t + 1)^2 dt.

Integrating, we get:

(1/4)ln|2t + 1| - (19/4)(1/(2t + 1)) + C,

where C is the constant of integration.

(ii) To evaluate the integral ∫(3/20)sin(x) dx, we can use the formula for the integral of sin(x):

∫sin(x) dx = -cos(x) + C,

where C is the constant of integration.

Multiplying by the coefficient, we have:

∫(3/20)sin(x) dx = -(3/20)cos(x) + C.

(b) To find dy/dx for the equation x + xy + xe = 4, we need to differentiate both sides of the equation with respect to x.

Differentiating x with respect to x gives 1.

To differentiate xy with respect to x, we need to use the product rule. The derivative of x with respect to x is 1, and the derivative of y with respect to x is dy/dx. So, the derivative of xy with respect to x is y + x(dy/dx).

To differentiate xe with respect to x, we can use the chain rule. The derivative of e^x with respect to x is e^x, and the derivative of x with respect to x is 1. So, the derivative of xe with respect to x is e^x + x(1).

Putting it all together, we have 1 + y + x(dy/dx) + e^x + x = 0.

Simplifying, we get dy/dx = -(1 + y + e^x)/(x + 1).

(a) To find all the roots α, β, γ of the cubic equation 3x^3 + 7x^2 - 6 = 0, we can use the factor theorem and synthetic division.

First, we check for any rational roots using the rational root theorem. The possible rational roots are factors of the constant term -6 divided by factors of the leading coefficient 3. In this case, the possible rational roots are ±1, ±2, ±3, ±6.

By trying these values, we find that x = -1 is a root of the equation.

Using synthetic division, we divide the cubic equation by (x + 1):

(x + 1) | 3 7 0 - 6 -3 -4 4 __________ 3 4 -4 0

The resulting quadratic equation is 3x^2 + 4x - 4 = 0.

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac))/(2a),

where a = 3, b = 4, and c = -4.

Calculating the values, we find that the roots of the quadratic equation are x = (-2 ± √10)/3.

Therefore, the roots of the cubic equation are α = -1, β = (-2 + √10)/3, and γ = (-2 - √10)/3.

To find the equation whose roots are α + β, β + γ, and α + γ, we can use Vieta's formulas.

The sum of the roots is α + β + γ = -1 + (-2 + √10)/3 + (-2 - √10)/3 = -5/3.

The product of the roots is αβγ = -1((-2 + √10)/3)((-2 - √10)/3) = (4 - 10)/9 = -6/9 = -2/3.

Therefore, the equation whose roots are α + β, β + γ, and α + γ is x^3 + (5/3)x^2 - (2/3)x + 2/3 = 0.