Compute the requested derivative. Do not simplify your answer after applying the differentiation rules.

Question

b) y=(x-3)/(x+4); Find d²y/dx²

c) y³+x³-5x(y-1)=2; Find dy/dx at (1,1)

For b, I tried to do implicit differentiation by first using the quotient rule on d2y/dx2, but I got stuck there.

For c, I got dy/dx=-2dx/3 but I don't think that's correct.

Answer 1

y = (x-3) / (x+4)

dy/dx = [(x+4)(1) - (x-3)(1)] / (x+4)² = 7/(x+4)² = 7(x+4)^-2

d²y/dx² = -14(x+4)^-3 = -14 / (x+4)³

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y³ + x³ - 5xy + 5x = 2

3y²(dy/dx) + 3x² - 5(y + x(dy/dx)) + 5 = 0

(dy/dx)(3y² - 5x) = -3x² + 5y - 5

dy/dx = (-3x² + 5y - 5) / (3y² - 5x)

Answer 2

b) Using the quotient rule,

dy/dx = [1(x + 4) - 1(x - 3)]/(x + 4)²

= (x + 4 - x + 3)/(x + 4)²

= 7/(x + 4)²

The first derivative can be rewritten as dy/dx=7(x + 4)^-2

Then d²y/dx² = -14(x + 4)^-3 = -14/(x + 4)³

c) Using implicit differentiation,

3y²(dy/dx) + 3x² - [5(y - 1) + 5x(dy/dx)] = 0

Note that the product rule was used inside the brackets in the previous step.

Now we substitute the point (1, 1) and solve for dy/dx:

3(1)²(dy/dx) + 3(1)² - [5(1 - 1) + 5(1)(dy/dx)] = 0

3(dy/dx) + 3 - 5(dy/dx) = 0

-2(dy/dx) = -3

dy/dx = 3/2