Question

If (x+y)^m+n = x^my^n .Then find dy/dx.

Answer 1

Given,

(x+y)^m+n = x^m*y^n .

Applying logarithm on both sides.

log (x+y)^m+n = logx^m*y^n .

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

(m+n) log ( x+y) = m logx + n logy .

Now, Differentiate with respect to " x "

d/dx [ (m+n) log(x+y) ] = d/dx [ mlogx + n logy ]

( m+n) (1/x+y) d/dx [ (x+y) ] = m (1/x) + n(1/y) dydx

(m+n)/x+y [ 1 + dy/dx ] = m/x + n/y dy/dx

[ (m+n)/(x+y) + (m+n)/(x+y) dy/dx ] = m/x + n/y dy/dx

dy/dx [ (m+n)/(x+y) - n/y ] = m/x -( m+n)/x+y
$\frac{dy}{dx} =\frac{ \frac{m(x + y) - (m + n)x}{(x + y)x} }{ \frac{(m + n)y - (x + y)n}{(x + y)y} } \\ \\ = \frac{ \frac{mx + my - mx - nx}{x} }{ \frac{my + ny - nx - ny}{y} } \\ \\ = \frac{my - nx}{x} \times \frac{y}{my - nx} \\ = \frac{y}{x}$

Formulae :

d/dx(u+v) = d/dx(u)+d/dx(v)
d/dx (Ku) = K d/dx ( u)
d/dx ( logx) = 1/x.

Hope helped!

Answer 2

Hi 

(x + y)^(m + n) = x^m . y^n 

takin natural log on both sides 

log(e){ (x + y)^(m + n) } = log(e) { x^m . y^n } 

(m + n) log(e) (x + y) = log(e) x^m + log(e) y^n

(m + n) log(e) (x + y) = m log(e) x  + n log(e) y 

Differentiate : 

(m + n) . 1/(x +y) .(1 + y') = m/x + n/y . y'

(m + n) .( 1 + y' / x+y)  = my + nxy' / xy 

(m + n) ( 1 + y' ) xy = ( my + nxy' ) (x + y ) 

mxy + mxyy' + nxy + nxyy' = mxy + nx²y + my² + nxyy' 

mxyy' + nxy = nx²y' + my²

mxyy' - nx²y = my² - nxy 

( my - nx ) xy' = ( my - nx ) y 

xy' = y 

y' = y/x