I really need help with this derivatives problem (calculus 1). (chegg tutors have given me very broad explanations so I've been stuck on this one for hours)

dy/dx = (1 − 8xy + 2y² + 16x⁴y² + 32x⁴y⁴) / (4x² − 4xy − 64x⁵y³)

Step-by-step explanation:

tan⁻¹(4x²y) = x + 2xy²

Take derivative of both sides with respect to x.  You'll need to use the following derivatives and rules:

Derivative of tan⁻¹: d/dx (tan⁻¹ u) = 1 / (1 + u²) du/dx

Product rule: d/dx (uv) = u dv/dx + v du/dx

Power rule: d/dx (u^n) = n u^(n−1) du/dx

(Notice the chain rule in each.)

So the derivative of the left side is:

d/dx [tan⁻¹(4x²y)] = 1 / (1 + (4x²y)²) × (4 (x² dy/dx + 2xy))

d/dx [tan⁻¹(4x²y)] = (4x² dy/dx + 8xy) / (1 + 16x⁴y²)

And the derivative of the right side is:

d/dx (x + 2xy²) = 1 + 2 (2xy dy/dx + y²)

d/dx (x + 2xy²) = 1 + 4xy dy/dx + 2y²

Set the sides equal and solve for dy/dx.

(4x² dy/dx + 8xy) / (1 + 16x⁴y²) = 1 + 4xy dy/dx + 2y²

4x² dy/dx + 8xy = (1 + 4xy dy/dx + 2y²) (1 + 16x⁴y²)

4x² dy/dx + 8xy = (1 + 2y²) (1 + 16x⁴y²) + 4xy (1 + 16x⁴y²) dy/dx

4x² dy/dx + 8xy = (1 + 2y² + 16x⁴y² + 32x⁴y⁴) + (4xy + 64x⁵y³) dy/dx

(4x² − 4xy − 64x⁵y³) dy/dx = 1 − 8xy + 2y² + 16x⁴y² + 32x⁴y⁴

dy/dx = (1 − 8xy + 2y² + 16x⁴y² + 32x⁴y⁴) / (4x² − 4xy − 64x⁵y³)

i think it would be 42cm

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!