Question

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

Answer 1

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.

Answer 2

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!

Answer 3

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`

Answer 4

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.