1.If 2xy+3yx=202x^y + 3y^x = 202xy+3yx=20, then dydx\frac{dy}{dx}dxdy at (2, 2) is equal to:a.−(2+loge83+loge4)- \left( \frac{2+\log_e 8}{3+\log_e 4} \right)−(3+loge42+loge8)b.−(3+loge164+loge8)- \left( \frac{3+\log_e 16}{4+\log_e 8} \right)−(4+loge83+loge16)c.−(3+loge82+loge4)- \left( \frac{3+\log_e 8}{2+\log_e 4} \right)−(2+loge43+loge8)d.−(3+loge42+loge8)- \left( \frac{3+\log_e 4}{2+\log_e 8} \right)−(2+loge83+loge4)Login to continueOnly logged in users canattempt or see the solution.
