If x^m × y^n=(x+y)^(m+n) find dy/dx?

1 Answer
Sep 3, 2017

#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