Compute dy/dx for the following curves: i) x^4 + y^4 = 10 ii)e^x/y = x^2 - y^2 i

Question

Compute dy/dx for the following curves:

i) x^4 + y^4 = 10

ii)e^x/y = x^2 - y^2

iii) x^2 + 4xy - y^2 = 4

iv) x^2 + 2xy - y^2 + x = 2 at (1,2)

Differentiation of Piece-wise Funtions:

Let F(x) = 1-x^2 if -5 < x <= 0

x^2 + 1 if - <= x < 5

Squareroot(x+2) if 5< x < infinity

i) f '(0) -

ii) f '(0) +

iii) f '(5) -

iv) f '(5) +

v) Is f(x) differentiable at x= 0?

vi) if f(x) differentiaable at x = 5?

For what values of x does the graph of f(x) x^3+3x^2+ x + 3 have a horizontal Tangent?

Explanation / Answer

i) x^4 + y^4 = 10 differentiate 4x^3 + 4y^3 dy/dx = 0 ==> dy/dx = -x^3/y^3 ii)e^x/y = x^2 - y^2 differentiate e^(x/y) * [ 1/y - x/y^2 * dy/dx] = 2x - 2y*dy/dx ==> e^(x/y) * [ y - x * dy/dx] = 2xy^2 - 2y^3*dy/dx ==> dy/dx = [2xy^2 - ye^(x/y)]/[2y^3 - xe^x/y)] iii) x^2 + 4xy - y^2 = 4 dofferentiate 2x + 4y + 4xdy/dx - 2ydy/dx = 0 ==> dy/dx = [-x - 2y]/[2x - y] iv) x^2 + 2xy - y^2 + x = 2 at (1,2) differentiate 2x + 2y + 2x dy/dx - 2y dy/dx + 1 = 0 put x = 1, y = 2 2 + 4 + 2dy/dx - 4dy/dx + 1 = 0 ==> dy/dx = 7/2 i) f '(0) - = -2x = -2*0 = 0 ii) f '(0) + = 2x = 2*0 = 0 iii) f '(5) - = 2x = 2*5 = 10 iv) f '(5) + = 1/[2*Sqrt(x+2)] = 1/[2*sqrt(7)] v) Is f(x) differentiable at x= 0? No. vi) if f(x) differentiaable at x = 5? No. f(x) = x^3+3x^2+ x + 3 differentiate f'(x) = 3x^2 + 6x + 1 horizontal tangent slope = 0 ==> f'(x) = 0 ==> 3x^2 + 6x + 1 = 0 ==> x = [-6 + sqrt(36 - 12)]/6, [-6 - sqrt(36 - 12)]/6 = [-3 + sqrt(6)]/3 , [-3 - sqrt(6)]/3

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