How do you implicitly differentiate #(cos(x^2+y^2))=(e^xy) #?

Question

3521 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-02T15:33:41

#y'=(-y*e^x-2x*sin(x^2+y^2))/(e^x+2y*sin(x^2+y^2))#

the given equation is
#cos(x^2+y^2)=e^x*y#

#d/dxcos(x^2+y^2)=d/dx(e^x*y)#

#-sin(x^2+y^2)*d/dx(x^2+y^2)=e^x*d/dx(y)+y*d/dx(e^x)#

#-sin(x^2+y^2)*(2x^(2-1)+2y^(2-1)*y')=e^x(y')+y(e^x)#

#-y*e^x-2x*sin(x^2+y^2)=e^x*y'+2y*sin(x^2+y^2)*y'#

#-y*e^x-2x*sin(x^2+y^2)=(e^x+2y*sin(x^2+y^2))*y'#

#(-y*e^x-2x*sin(x^2+y^2))/(e^x+2y*sin(x^2+y^2))=((e^x+2y*sin(x^2+y^2))*y')/((e^x+2y*sin(x^2+y^2))#

#(-y*e^x-2x*sin(x^2+y^2))/(e^x+2y*sin(x^2+y^2))=cancel((e^x+2y*sin(x^2+y^2))*y')/(cancel(e^x+2y*sin(x^2+y^2))#

#y'=(-y*e^x-2x*sin(x^2+y^2))/(e^x+2y*sin(x^2+y^2))#

God bless....I hope the explanation is useful.