Given that #xe^y = x+1#, show that #e^y + xe^ydy/dx = 1#?

4 Answers
May 12, 2018

Given: #xe^y = x+1#

Begin the implicit differentiation process:

#(d(xe^y))/dx = dx/dx+(d1)/dx#

The derivative of a constant is 0:

#(d(xe^y))/dx = dx/dx#

#dx/dx# becomes 1:

#(d(xe^y))/dx = 1" [1]"#

When attempting to compute a derivative of the form #(d(uv))/dx#, the product rule states that

#(d(uv))/dx = (du)/dxv+u(dv)/dx#

The left side of equation [1] is in this form where #u = x# and #v = e^y#

Substituting this into the product rule we obtain the equation:

#(d(xe^y))/dx = (d(x))/dx e^y+ x(d(e^y))/dx#

We know that #(d(x))/dx = 1#:

#(d(xe^y))/dx = e^y+ x(d(e^y))/dx#

#e^y# is a function of y, therefore, we must use the chain rule to compute #(d(e^y))/dx# as follows:

#(d(e^y))/dx = (d(e^y))/dydy/dx#

The exponential function is a special case where #(d(e^y))/dy = e^y#:

#(d(e^y))/dx = e^ydy/dx#

After using the product rule and the chain rule, we have#(d(xe^y))/dx = e^y+ xe^ydy/dx# and we substitute this into equation [1]:

#e^y+ xe^ydy/dx = 1#

We shall stop here because the above is what we want.

May 12, 2018

We use implicit differentiation to see that

#e^y + xe^y(dy/dx) = 1#

#xe^y(dy/dx) = 1 -e^y#

#dy/dx= (1 - e^y)/(xe^y)#

We now substitute into the given expression.

#e^y + xe^y(1 - e^y)/(xe^y) = 1#

#e^y+ 1 - e^y = 1#

#1 = 1#

As required.

Hopefully this helps!

May 12, 2018

Please see the explanation below

Explanation:

Another way of proving this is :

Let,

#f(x,y)=xe^y-x-1#

And

#dy/dx=-((delf) /(delx))/((delf) /(dely))#

#(delf) /(delx)=e^y-1#

#(delf) /(dely)=xe^y#

Therefore,

#dy/dx=-(e^y-1)/(xe^y)#

Rearranging,

#xe^ydy/dx=-e^y+1#

#e^y+xe^ydy/dx=1#

May 12, 2018

Given expression:

# x e^y = x + 1 #

Inspection reveals that differential of RHS with respect to #x# gives us RHS of the expression required to be proved. Therefore, differentiating both sides with respect to variable #x# we get

# d/(dx)( x e^y) =d/(dx) (x + 1) #
# d/(dx)( x e^y) =1#

Using product rule and implicit differentiation on the LHS we get

#LHS= d/(dx)( x e^y)#
#=>LHS= x d/(dx) e^y+e^yd/(dx)x#
#=>LHS= x e^y(dy)/(dx)+e^y#

We see that it is same as #LHS# of the expression which was to be proved.

Hence Proved.