Question

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

Answer 1

`\text{Given that,}\\\\~~~1+ \ln (3xy) = e^(3x-y)\\\\\\\implies \frac d{dx} \left[1 + \ln (3xy) \right] = (d)/(dx) \left(e^(3x-y) \right)\\\\\\\implies (1)/(3xy) \cdot 3 (d)/(dx)(xy) = e^(3x-y) \cdot \left(3 - (dy)/(dx) \right)\\ \\\\\implies \frac 1{xy} \left( x (dy)/(dx) + y \right)= e^(3x-y) \cdot \left(3 - (dy)/(dx) \right)\\\\\\\implies \frac 1y (dy)/(dx) + \frac 1x = 3e^(3x-y)- e^(3x -y) (dy)/(dx)`

`\implies \frac 1y (dy)/(dx) + e^(3x-y) (dy)/(dx) = 3e^(3x-y) - \frac 1x\\\\\\\implies \left( \frac 1y+ e^(3x-y) \right)(dy)/(dx) = 3e^(3x-y) - \frac 1x\\\\\\\implies (dy)/(dx)=(3 e^(3x-y) -\frac 1x)/(\frac 1y+ e^(3x-y) )\\\\\text{At point}~ (x,y) = (1/3, 1), \text{slope of the tangent line,}\\\\m=(dy)/(dx) = 0\\\\\text{Hence the equation of tangent line,}\\\\~~~y-1 =0\\\\\implies y=1`