Can someone help me find the derivatives of the following problems?

Question

a.(x^2 + y^2)2 = 2x^2 - 2y^2
b. 2x^2 +4xy^2=sinxy +8
c. (ln^2x)/e^(3x+1)
and can you explain how you got to your answers?

Answer 1

(a).

The equation is (x² + y²)² = 2x² - 2y².

Differentiate with respect to x.

(d/dx)(x² + y²)² = (d/dx)(2x² - 2y²).

2(x² + y²)[2x+2y*y'] = 4x - 4y*y'

x(x² + y²)+(x² + y²)y*y' = x - y*y'

(x² + y²)y*y' + y*y' = x - x(x² + y²)

[(x² + y²)y + y]*y' = x - x(x² + y²)

y' = [x - x(x² + y²)]/[(x² + y²)y + y]

Answer 2

(b).

The equation is 2x² + 4xy² = sin(xy) + 8.

Differentiate with respect to x.

(d/dx)(2x² + 4xy²) = (d/dx)(sin(xy) + 8).

4x + 4[x*2y*y' + y²) = cos(xy)*[x*y' + y]

4x + 8xyy' + 4y² = xy'cos(xy) + ycos(xy)

8xyy' - xy'cos(xy) = ycos(xy) - 4y² - 4x

y'[8xy - xcos(xy)] = ycos(xy) - 4y² - 4x

y' = [ycos(xy) - 4y² - 4x]/[8xy - xcos(xy)]

Answer 3

(c).

The expression is [ln(x)]²/e^3x+1.

Differentiate with respect to x.

= (d/dx)([ln(x)]²/e^3x+1)

Apply quotient rule of derivative: (u/v) = [vu' - uv']/v².

= [e^3x+1*2ln(x)*(1/x) - [ln(x)]²*e^3x+1*3]/(e^3x+1)²

= [2ln(x)*(1/x) - 3[ln(x)]²]/(e^3x+1)

= [2ln(x) - 3x[ln(x)]²]/x(e^3x+1)

(d/dx)([ln(x)]²/e^3x+1) = [2ln(x) - 3x[ln(x)]²]/x(e^3x+1).