Find \tt \: (dy)/(dx) when \tt {x}^(2) + {y}^(2) = log(x + y) Please help!​

Question

Find
`\tt \: (dy)/(dx)` when
`\tt {x}^(2) + {y}^(2) = log(x + y)`

Please help!​

Answer 1

`\boxed{\bold{\tt(dy)/(dx)= (2x^2+2xy-1)/(1-2xy-2y^2)}}`

Explanation:

x^2 + y^2 = log(x+y)

Differentiating both sides with respect to x.

`\bold{\tt (d)/(dx) (x^2 + y^2) =(d)/(dx) log(x+y)}`

`\bold{\tt{Apply\:the\:Sum/Difference\:Rule}:}`

`\bold{\tt (d)/(dx)\left(x^2\right)+(d)/(dx)\left(y^2\right) =(d)/(dx) log(x+y)}`

Apply Power rule and chain rule

`\bold{\tt2x+2y(dy)/(dx)=(d)/(dx) log(x+y)}`

`\tt Apply\:the\:chain\:rule:`

`\bold{\tt 2x+2y(dy)/(dx)= (1)/(x+y)(d)/(dx)\left(x+y\right)}`

`\bold{\tt{Apply\:the\:Sum/Difference\:Rule}:}`

`\bold{\tt 2x+2y(dy)/(dx)=(1)/(x+y)(d)/(dx)x+(1)/(x+y)(d)/(dx)*y}`

`\bold{\tt 2x+2y(dy)/(dx)=(1)/(x+y)(d)/(dx)x+(1)/(x+y)(d)/(dy)*y*(dy)/(dx)}`

`\bold{\tt 2x+2y(dy)/(dx)=(1)/(x+y)+(1)/(x+y)(dy)/(dx)}`

Solving for
`\tt (dy)/(dx)`

`\bold{\tt 2x-(1)/(x+y)=(1)/(x+y)(dy)/(dx)-2y(dy)/(dx)}`

`\bold{\tt(1)/(x+y)(dy)/(dx)-2y(dy)/(dx)= 2x-(1)/(x+y)}`

`\bold{\tt(dy)/(dx)((1)/(x+y)-2y)= 2x-(1)/(x+y)}`

`\bold{\tt(dy)/(dx)((1-2y(x+y))/(x+y))= (2x(x+y)-1)/(x+y)}`

`\bold{\tt(dy)/(dx)= ((2x(x+y)-1)/(x+y))/(((1-2y(x+y))/(x+y)))}`

`\bold{\tt(dy)/(dx)= (2x(x+y)-1)/(1-2y(x+y))}`

`\bold{\tt(dy)/(dx)= (2x^2+2xy-1)/(1-2xy-2y^2)}`

Therefore,Answer is:
`\boxed{\bold{\tt(dy)/(dx)= (2x^2+2xy-1)/(1-2xy-2y^2)}}`

Note: Formula

`\boxed{\bold{\tt{Addition \: Rule:(d)/(dx)(x^n+y^n) =(d)/(dx)*x^n+(d)/(dx)*y^n}}}`

`\boxed{\bold{\tt{Power \: Rule:(d)/(dx)x^n =n*x^(n-1)}}}`

`\boxed{\bold{\tt{Chain \:\: Rule: (d)/(dx)y^n=(d)/(dy)y^n(dy)/(dx)=n*y^(n-1)(dy)/(dx)}}}`

`\boxed{\bold{\tt{Product\:Rule:(d)/(dx)(u*v)=(du)/(dx)*v+u*(dv)/(dx)}}}`

Answer 2

`\frac{\text{d}y}{\text{d}x}=(1-2x^2-2xy)/(2xy+2y^2-1)`

Explanation:

Given equation:

`x^2+y^2=\log(x+y)`

Assuming log(x + y) is the natural log:

`x^2+y^2=\ln(x+y)`

To differentiate an equation that contains a mixture of x and y terms, use implicit differentiation.

Begin by placing d/dx in front of each term of the equation:

`\frac{\text{d}}{\text{d}x}x^2+\frac{\text{d}}{\text{d}x}y^2=\frac{\text{d}}{\text{d}x}\ln(x+y)`

Differentiate the left side of the equation first.

`\boxed{\begin{minipage}{4.5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\frac{\text{d}y}{\text{d}x}=nx^(n-1)$\\\end{minipage}}`

Differentiate the terms in x only using the above rule:

`2x+\frac{\text{d}}{\text{d}x}y^2=\frac{\text{d}}{\text{d}x}\ln(x+y)`

Use the chain rule to differentiate terms in y only.

In practice, this means differentiate with respect to y, and place dy/dx at the end:

`2x+2y\frac{\text{d}y}{\text{d}x}=\frac{\text{d}}{\text{d}x}\ln(x+y)`

Now we have differentiated the left side of the equation, we can differentiate the right side of the equation.

`\boxed{\begin{minipage}{6 cm}\underline{Differentiating $\ln(f(x))$}\\\\If $y=\ln(f(x))$, then $\frac{\text{d}y}{\text{d}x}=(1)/(f(x))\cdot f'(x)$\\\end{minipage}}`

Apply the rule to differentiate ln(x + y):

`2x+2y\frac{\text{d}y}{\text{d}x}=(1)/(x+y)\cdot \frac{\text{d}}{\text{d}x}(x+y)`

Differentiate (x + y):

`2x+2y\frac{\text{d}y}{\text{d}x}=(1)/(x+y)\left(1+\frac{\text{d}y}{\text{d}x}\right)`

Simplify:

`2x+2y\frac{\text{d}y}{\text{d}x}=(1)/(x+y)+(1)/(x+y)\frac{\text{d}y}{\text{d}x}\right)`

Rearrange the resulting equation to isolate dy/dx:

`2y\frac{\text{d}y}{\text{d}x}-(1)/(x+y)\frac{\text{d}y}{\text{d}x}\right)=(1)/(x+y)-2x`

`\left(2y-(1)/(x+y)\right)\frac{\text{d}y}{\text{d}x}=(1)/(x+y)-2x`

`\left((2y(x+y))/(x+y)-(1)/(x+y)\right)\frac{\text{d}y}{\text{d}x}=(1)/(x+y)-(2x(x+y))/(x+y)`

`\left((2y(x+y)-1)/(x+y)\right)\frac{\text{d}y}{\text{d}x}=(1-2x(x+y))/(x+y)`

`\frac{\text{d}y}{\text{d}x}=(1-2x(x+y))/(x+y) / (2y(x+y)-1)/(x+y)`

`\frac{\text{d}y}{\text{d}x}=(1-2x(x+y))/(x+y) \cdot (x+y)/(2y(x+y)-1)`

`\frac{\text{d}y}{\text{d}x}=(1-2x(x+y))/(2y(x+y)-1)`

To simplify further, expand the brackets:

`\frac{\text{d}y}{\text{d}x}=(1-2x^2-2xy)/(2xy+2y^2-1)`