d/dx x(1+y)^(1/2) + y(1+x)^(1/2)=0. Can some1 show me the calculation because i didn't get like the answer.
[ans=-1/(1+x)^2 but my calculation=-(sqrt(y+1) (2 sqrt(x+1) sqrt(y+1)+y)/(sqrt(x+1) (2 sqrt(x+1) sqrt(y+1)+x)]
To find the derivative of the given expression with respect to x, we can use the chain rule. The chain rule states that if we have a function f(g(x)), then the derivative is given by f'(g(x)) * g'(x).
Let's break down the given expression and apply the chain rule step-by-step:
d/dx [ x√(1+y) + y√(1+x) ] = 0
First, let's differentiate the first term: x√(1+y), with respect to x.
Using the chain rule, the derivative of √(1+y) with respect to x is:
d/dx √(1+y) = (1/2) * (1+y)^(-1/2) * dy/dx
Considering 'y' as a function of x, we can write dy/dx as y'.
So, the first term becomes: x * (1/2) * (1+y)^(-1/2) * y'
Moving on to the second term: y√(1+x).
Again, using the chain rule, the derivative of √(1+x) with respect to x is:
d/dx √(1+x) = (1/2) * (1+x)^(-1/2) * dx/dx
Since dx/dx is simply 1, we can simplify:
d/dx √(1+x) = (1/2) * (1+x)^(-1/2)
Now, let's calculate the derivative of the second term:
= y * (1/2) * (1+x)^(-1/2) * 1
Combining the derivative of the first term and the derivative of the second term, we have:
x * (1/2) * (1+y)^(-1/2) * y' + y * (1/2) * (1+x)^(-1/2) = 0
To solve for y', we can isolate it:
x * (1/2) * (1+y)^(-1/2) * y' = - y * (1/2) * (1+x)^(-1/2)
Now, divide both sides of the equation by x * (1/2) * (1+y)^(-1/2):
y' = - y * (1/2) * (1+x)^(-1/2) / (x * (1/2) * (1+y)^(-1/2))
Simplifying further:
y' = - y * (1+y)^(-1/2) / [x * (1+x)^(-1/2)]
To simplify the expression, we can multiply the numerator and denominator by (1+y)^(1/2) * (1+x)^(1/2):
y' = - y * (1+y)^(-1/2) * (1+y)^(1/2) * (1+x)^(1/2) / [x * (1+x)^(-1/2) * (1+y)^(1/2) * (1+x)^(1/2)]
Simplifying further:
y' = - y * (1+x) / [x * (1+x)^(-1)] = - y * (1+x) / (x * (1+x))
Now, let's simplify the expression:
y' = -y / x
Therefore, the correct derivative is: y' = -y / x, which is equivalent to -1 / (1+x)^2.
So, the answer is indeed -1 / (1+x)^2.