Calculus question...hard one?
ln(x+y)=tan^-1(y)
find dy/dx.
this kind of question I am suppose to use implicit differentiation, please help me guys.
Derivative of the left side:
d( ln(x+y) )/dx
1/(x+y) *d(x+y)/x
1(x+y) * (1 + dy/dx)
Derivative of the right side:
d( arctan(y) )/dx
1/(y^2+1) * d(y)/dx
Set these two equal and perform algebraic manipulations to solve for dy/dx:
1(x+y) * (1 + dy/dx) = 1/(y^2+1) * d(y)/dx
...
dy/dx = (y^2 + 1) / (x + y - y^2 - 1)
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Verified answer
Derivative of the left side:
d( ln(x+y) )/dx
1/(x+y) *d(x+y)/x
1(x+y) * (1 + dy/dx)
Derivative of the right side:
d( arctan(y) )/dx
1/(y^2+1) * d(y)/dx
Set these two equal and perform algebraic manipulations to solve for dy/dx:
1(x+y) * (1 + dy/dx) = 1/(y^2+1) * d(y)/dx
...
...
...
dy/dx = (y^2 + 1) / (x + y - y^2 - 1)