Question

50 points each question. Please help. How do I solve?

Answer 1

`~~~~~~\tan^(-1)(xy) = \sin^(-1)(4x+4y)\\\\\\\implies \frac d{dx} \left[ \tan^(-1)(xy)\right] = \frac d{dx} \sin^(-1)\left( 4x+4y\right)\\ \\\\\implies (1)/(1+(xy)^2) \left[ (d)/(dx) (xy)\right] = (1)/(√(1-(4x+4y)^2)) \cdot (d)/(dx)(4x+4y)\\\\\\\implies (1)/(1+x^2 y^2) \left(x(dy)/(dx)+y \right)=(1)/(√(1-(4x+4y)^2)) \left(4+ 4 (dy)/(dx) \right)\\\\\\\\`

`\implies \left((x)/(1+x^2y^2) \right) (dy)/(dx) + (y)/(1+x^2y^2)=(1)/(√(1-(4x+4y)^2)) \left(4+ 4 (dy)/(dx) \right)\\`

`\\\\\implies 0 \cdot (dy)/(dx) +0 = (1)/(√(1 -0)) \left(4+4 (dy)/(dx) \right) ~~~~~~~~~;\left[ \text{at}~~ (0,0) \right]\\\\\\\implies 4+4 (dy)/(dx) = 0\\\\\\\implies (dy)/(dx) = -\frac 44\\\\\\\implies (dy)/(dx) = -1\\\\\text{So, the slope of the tangent line at (0,0) is}~ -1\\ \\\\\text{The equation of tangent line,}\\\\~~~~~~y-0 = -1(x-0)\\\\\implies y=-x`