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

2 Answers
Jan 25, 2018

#dy/dx=(e^(xy)-(x-y^2)e^(xy)y)/(1+2y-(x-y^2)e^(xy)x)#

Explanation:

#y=(x-y^2)e^(xy)#

We could proceed using the product rule. In implicit differentiation, we must also remember to use the chain rule when differentiating a #y# since #y# is a function of #x#. So:

#y=(x-y^2)e^(xy)#

#dy/dx=d/dx(x-y^2)*e^(xy)+(x-y^2)d/dxe^(xy)#

#dy/dx=(1-2ydy/dx)e^(xy)+(x-y^2)*e^(xy)*d/dx{xy}#

#dy/dx=(1-2ydy/dx)e^(xy)+(x-y^2)*e^(xy)(y+xdy/dx)#

Now gather all terms with #dy/dx# on the left side:

#dy/dx+2ydy/dx-(x-y^2)e^(xy)xdy/dx=e^(xy)+(x-y^2)e^(xy)y#

#dy/dx(1+2y-(x-y^2)e^(xy)x)=e^(xy)+(x-y^2)e^(xy)y#

#->dy/dx=(e^(xy)-(x-y^2)e^(xy)y)/(1+2y-(x-y^2)e^(xy)x)#

Jan 25, 2018

#dy/dx=(xye^(xy)-y^3e^(xy)+e^(xy))/(1-x^2e^(xy)+xy^2e^(xy)+2ye^(xy))#

Explanation:

#"differentiate using the "color(blue)"product/chain rules"#

#"given "y=f(x)g(x)" then"#

#dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"#

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#f(x)=x-y^2rarrf'(x)=1-2ydy/dx#

#g(x)=e^(xy)rArrg'(x)=e^(xy)xxd/dx(xy)#

#color(white)(xxxxxxxxxxxxx)=e^(xy)(xdy/dx+y)#

#color(white)(xxxxxxxxxxxxx)=xe^(xy)dy/dx+ye^(xy)#

#(x-y^2)(xe^(xy)dy/dx+ye^(xy)))+e^(xy)(1-2ydy/dx)#

#=x^2e^(xy)dy/dx+xye^(xy)-xy^2e^(xy)dy/dx-y^3e^(xy)+e^(xy)-2ye^(xy)dy/dx#

#y=(x-y^2)e^(xy)larrcolor(blue)"original question"#

#rArrdy/dx-x^2e^(xy)dy/dx+xy^2e^(xy)dy/dx+2ye^(xy)dy/dx=xye^(xy)-y^3e^(xy)+e^(xy)#

#rArrdy/dx(1-x^2e^(xy)+xy^2e^(xy)+2ye^(xy))=xye^(xy)-y^3e^(xy)+e^(xy)#

#rArrdy/dx=(xye^(xy)-y^3e^(xy)+e^(xy))/(1-x^2e^(xy)+xy^2e^(xy)+2ye^(xy))#