How do you implicitly differentiate #-1=(x+y)^2-xy-e^(3x+7y) #?

Question

How do you implicitly differentiate #-1=(x+y)^2-xy-e^(3x+7y)  #?

Isaiah Allen · Accepted Answer

#y'={2x+y-3(x^2+y^2+xy+1)}/{7(x^2+y^2+xy+1)-(2y+x)}.#

We will use the given eqn. as #e^(3x+7y)=(x+y)^2-xy+1=x^2+y^2+xy+1,# & diff. its both sides, to get,

#e^(3x+7y)*(3+7y')=2x+2yy'+xy'+y.# Subbing #x^2+y^2+xy+1# for #e^(3x+7y)# in #L.H.S.,#

#3(x^2+y^2+xy+1)+7y'(x^2+y^2+xy+1)=2x+y+(2y+x)y'#

#y'{7(x^2+y^2+xy+1)-(2y+x)}=2x+y-3(x^2+y^2+xy+1)#

#y'={2x+y-3(x^2+y^2+xy+1)}/{7(x^2+y^2+xy+1)-(2y+x)}.#

I find this soln. of mine very easier than the one I provided earlier!

Charles Anctil · Answer

#y'={2x+y-3e^(3x+4y)}/(4e^(3x+4y)-x-2y).#

Given Eqn. is #-1=(x+y)^2-xy-e^(3x+7y),# or, #xy+e^(3x+4y)=(x+y)^2+1.#

Diff. both sides,

#(xy)'+{e^(3x+\color{red}{7}y)}'={(x+y)^2}'+0.# #:. xy'+yx'+e^(3x+7y)(3x+7y)'=2(x+y)(x+y)'# #:.xy'+y+e^(3x+7y){(3x)'+(7y)'}=2(x+y)(x'+y')# #:. xy'+y+e^(3x+7y)(3+7y')=2(x+y)(1+y'),# i.e., #xy'+y+3e^(3x+4y)+7y'e^(3x+7y)=2(x+y)+2y'(x+y).# #:. xy'+7y'e^(3x+7y)-2y'(x+y)=2(x+y)-y-3e^(3x+4y).# #:. y'(x+7e^(3x+7y)-2x-2y)=2x+2y-y-3e^(3x+4y),# or, #y'(7e^(3x+7y)-x-2y)=2x+y-3e^(3x+4y).#

Hence, #y'={2x+y-3e^(3x+7y)}/(7e^(3x+7y)-x-2y).#

#y'# can further be simplified, as below :- For this, we write the given eqn. as #e^(3x+7y)=(x+y)^2+1-xy=x^2+xy+y^2+1,# & submit the value of e^(3x+7y)) in #y'# to give,

#y'={2x+y-3(x^2+xy+y^2+1)}/{4(x^2+xy+y^2+1)-x-2y}.#

Paisley Johnson · Answer

undefined To implicitly differentiate the given equation $ -1 = (x + y)^2 - xy - e^{3x + 7y} $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $ using the chain rule and product rule where necessary.
2. Express the derivative of $ y $ with respect to $ x $ as $ \frac{dy}{dx} $.
3. Solve for $ \frac{dy}{dx} $ to find the implicit derivative.

Differentiating the equation with respect to $ x $:

$$ \frac{d}{dx}(-1) = \frac{d}{dx}((x + y)^2 - xy - e^{3x + 7y}) $$

$$ 0 = 2(x + y) \frac{d}{dx}(x + y) - y - x \frac{dy}{dx} - e^{3x + 7y}(3 + 7\frac{dy}{dx}) $$

$$ 0 = 2(x + y)(1 + \frac{dy}{dx}) - y - x \frac{dy}{dx} - e^{3x + 7y}(3 + 7\frac{dy}{dx}) $$

$$ 0 = 2(x + y) + 2(x + y)\frac{dy}{dx} - y - x \frac{dy}{dx} - 3e^{3x + 7y} - 7e^{3x + 7y}\frac{dy}{dx} $$

$$ 0 = 2(x + y) + (2(x + y) - y - 3e^{3x + 7y})\frac{dy}{dx} - x \frac{dy}{dx} - 7e^{3x + 7y}\frac{dy}{dx} $$

$$ 0 = 2(x + y) + (2x + 2y - y - 3e^{3x + 7y})\frac{dy}{dx} - x \frac{dy}{dx} - 7e^{3x + 7y}\frac{dy}{dx} $$

$$ 0 = 2(x + y) + (2x + y - 3e^{3x + 7y})\frac{dy}{dx} - x \frac{dy}{dx} - 7e^{3x + 7y}\frac{dy}{dx} $$

$$ 0 = 2x + 2y - 3e^{3x + 7y} - x \frac{dy}{dx} - 7e^{3x + 7y}\frac{dy}{dx} $$

$$ (2y - 3e^{3x + 7y}) = (x + 7e^{3x + 7y})\frac{dy}{dx} $$

$$ \frac{dy}{dx} = \frac{2y - 3e^{3x + 7y}}{x + 7e^{3x + 7y}} $$

Therefore, the implicit derivative of $ y $ with respect to $ x $ is $ \frac{2y - 3e^{3x + 7y}}{x + 7e^{3x + 7y}} $.