Question

If x^m × y^n=(x+y)^(m+n) find dy/dx?

Answer 1

`dy/dx = y/x`

Explanation:

We have:

`x^m y^n = (x+y)^(m+n)`

Take (natural) logarithms of both sides:

`ln(x^m y^n) = ln((x+y)^(m+n))`

Then using the properties of logarithms we can simplify:

`ln(x^m) + ln( y^n) = ln((x+y)^(m+n))`
`:. m ln x + n ln y = (m+n)ln(x+y)`

Now we can differentiate implicitly:

`m/x + n/yy' = (m+n)/(x+y)(1+y')`

`:. m/x + n/yy' = (m+n)/(x+y) + (m+n)/(x+y) y'`

`:. (m+n)/(x+y) y' - n/yy' = m/x - (m+n)/(x+y)`

`:. ( (m+n)y - n(x+y))/( (x+y)y )y' = (m(x+y) - (m+n)x)/(x(x+y))`

`:. ( my+ny - nx-ny )/( (x+y)y )y' = ( mx+my - mx-nx )/(x(x+y))`

`:. ( my - nx )/( (x+y)y )y' = ( my -nx )/(x(x+y))`

`:. y' = y/x`

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find `y` explicitly as a function of `x`, only implicitly through the equation `F(x, y) = 0` which defines `y` as a function of `x, y = y(x)`. Therefore we can write `F(x, y) = 0` as `F(x, y(x)) = 0`. Differentiating both sides of this, using the partial chain rule gives us

`(partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))`

So Let `F(x,y) = x^m y^n - (x+y)^(m+n)`; Then;

`(partial F)/(partial x) = mx^(m-1)y^n - (m+n)(x+y)^(m+n-1)`

`(partial F)/(partial y) = nx^my^(n-1) - (m+n)(x+y)^(m+n-1)`

And so:

`dy/dx = - (mx^(m-1)y^n - (m+n)(x+y)^(m+n-1)) / (nx^my^(n-1) - (m+n)(x+y)^(m+n-1))`

`\ \ \ \ \ = - (mx^(m-1)y^n - (m+n)(x+y)^(m+n)(x+y)^(-1)) / (nx^my^(n-1) - (m+n)(x+y)^(m+n)(x+y)^(-1))`

`\ \ \ \ \ = - (mx^(m-1)y^n - (m+n)x^m y^n(x+y)^(-1)) / (nx^my^(n-1) - (m+n)x^m y^n(x+y)^(-1))`

`\ \ \ \ \ = - (m/x - (m+n)/(x+y)) / (n/y - (m+n)/(x+y))`

`\ \ \ \ \ = - ((m(x+y) - (m+n)x)/(x(x+y))) / ((n(x+y) - (m+n)y) / (y(x+y)))`

`\ \ \ \ \ = - ((mx+my -mx-nx)/(x)) / ((nx+ny - my-ny) / (y))`

`\ \ \ \ \ = - ((my -nx)/(x)) / ((nx- my) / (y))`
`\ \ \ \ \ = ((nx-my)/(x)) / ((nx- my) / (y))`
`\ \ \ \ \ = y/x`, as before