Differentiation help please?

Question

1. Given that xy = 8100, find the value of dy/dx when y = 180.
2. The gradient of the curve y = px + q/x^2 at the point (1,9) is 3. Calculate the values of p and q.
3. Differentiate (3t^2 - 4t)(t + 1)/2(square root t) with respect to t.

Answer 1

xy = 8100

Substitute y = 180 in the equation.

x(180) = 8100.

Divide each side by 180.

x = 8100/180 = 45.

Take xy = 8100.

Apply 'derivative with respect to x' each side.

(d/dx)(xy) = (d/dx)(8100)

The Product Rule (UV)' = U'V + UV' and derivative of constant is 0.

x'y + xy' = 0

(1)y + xy' = 0

Substitute x = 45 and y = 180 in the equation.

180 + 45(y') = 0

Subtract 180 from each side.

45(y') = -180

y' = -4

Therefore the value of (dy/dx) = -4.

Answer 2

3). (3t² - 4t)(t + 1)/2

FOIL method: the product of two binomials is the sum of the products of the First terms, the Outer terms, the Inner terms and the Last terms.

[(3t²)t + (3t²)1 - (4t)t - (4t)1]/2.

[3t³ + 3t² - 4t² - 4t]/2

[3t³ - t² - 4t]/2

Apply 'derivative with respect to t' each side.

[(d/dt)3t³ - (d/dt)t² - (d/dt)4t]/2

Power Rule of Derivative (d/dx)(xⁿ) = nx^n-1

[3(3t²) - (2t) - 4]/2

[9t² - 2t - 4]/2

9t²/2 - 2t/2 - 4/2

(9/2)t² - t - 2.

Answer 3

2). The gradient of the curve y = px + q/x² at the point (1,9) is 3

i.e. y' = 3 at (1, 9)

9 = p(1) + q/(1)²

9 = p + q -----------------> (1)

Take y = px + q/x²

Mathematical rule: 1/a = a^-1

y = px + qx^-2

Apply derivative each side.

dy/dx = (d/dx)(px + qx^-2)

y' = (d/dx)(px) + (d/dx)(qx^-2)

Power Rule of Derivative (d/dx)(xⁿ) = nx^n-1

y' = p(1) + q(-2x^-3)

y' = p - 2q/x³

Substitute x = 1 and y' = 3 in the equation.

3 = p - 2q/1³

3 = p - 2q------------>(2)

Subtract equation (1) and equation (2).

p + q = 9

p - 2q = 3

--------------------

3q = 6

Divide each side by 3.

q = 6/3 = 2

Substitute q = 2 in the equation (1)

p + 2 = 9

Subtract 2 from each side.

p = 9 - 2 = 7

Therefore p = 7 and q = 2.

Answer 4

=

Use product rule : .

Use power rule : .

.

Therefore, .