How do you implicitly differentiate #4= xytan(x^2y) #?

Question

1219 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-10T12:54:43

#y'=(-y* tan(x^2y)-2x^2y^2*sec^2(x^2 y))/(x* tan(x^2 y)+x^3 y* sec^2(x^2 y))#

from the given

#4=xy tan(x^2 y)#

differentiate both sides of the equation

#d/dx(4)=d/dx(x)*y*tan(x^2 y)+d/dx(y)*x*tan(x^2 y)+#

#xy*d/dx(tan(x^2 y)#

#0=y*tan(x^2 y)+x*tan(x^2 y) y' + 2x^2 y^2 *sec^2(x^2 y) + x^3 y##sec^2(x^2 y) y'#

#-y tan(x^2 y)-2 x^2 y^2 sec^2 (x^2 y) = (x tan (x^2 y) + x^3 y sec^2(x^2 y)) y'#

#y'=(-y* tan(x^2y)-2x^2y^2*sec^2(x^2 y))/(x* tan(x^2 y)+x^3 y* sec^2(x^2 y))#