How do you implicitly differentiate #2=(x+2y)^2-xy-e^(3x+y^2) #?

Question

How do you implicitly differentiate #2=(x+2y)^2-xy-e^(3x+y^2)  #?

Paisley Johnson · Accepted Answer

#y'=(2x+3y-3*e^((3x+y^2)))/(-3x-8y+2y*e^((3x+y^2)))#

From the given differentiate each term of both sides with respect to x

#2=(x+2y)^2-xy-e^(3x+y^2)#

#d/dx(2)=d/dx(x+2y)^2-d/dx(xy)-d/dx(e^(3x+y^2))#

#0=2(x+2y)^(2-1)*d/dx(x+2y)-xdy/dx-y*d/dx(x)-e^(3x+y^2)*d/dx(3x+y^2)#

#0=2(x+2y)^(1)*(1+2y')-xy'-y*1-e^(3x+y^2)*(3+2yy')#

#0=(2x+4y)(1+2y')-xy'-y-e^(3x+y^2)(3+2yy')#

Expand then simplify

#0=2x+4y+4xy'+8yy'-xy'-y-3e^(3x+y^2)-2yy'e^(3x+y^2)#

Transpose those terms with y' to the left of the equation

#-4xy'-8yy'+xy'+2yy'e^(3x+y^2)=2x+4y-y-3e^(3x+y^2)#

factor out the #y'#

#(-4x-8y+x+2ye^(3x+y^2))y'=2x+4y-y-3e^(3x+y^2)#

simplify

#(-3x-8y+2ye^(3x+y^2))y'=2x+3y-3e^(3x+y^2)#

divide both sides by #(-3x-8y+2ye^(3x+y^2))#

#(-3x-8y+2ye^(3x+y^2))y'=2x+3y-3e^(3x+y^2)#

#cancel((-3x-8y+2ye^(3x+y^2))y')/cancel((-3x-8y+2ye^(3x+y^2)))#

#=(2x+3y-3e^(3x+y^2))/(-3x-8y+2ye^(3x+y^2))#

and

#y'=(2x+3y-3e^(3x+y^2))/(-3x-8y+2ye^(3x+y^2))#

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

Luke Alvarez · Answer

undefined User is studying implicit differentiation.To implicitly differentiate $2=(x+2y)^2-xy-e^{3x+y^2}$, differentiate each term with respect to $x$ and simplify the result. The derivative of a constant is 0, so the derivative of 2 with respect to $x$ is 0. For the other terms, use the chain rule and product rule as needed. The result after simplification should give you the implicit derivative.