Solve f(x, y)=\frac{2 y}{y+\cos x}

Simplify

f (x, y) = \frac{2 y}{y + cos ( x )}

Find the first partial derivative by treating the variable y as a constant and differentiating with respect to x

f_{x} = \frac{\partial}{\partial x} (\frac{2 y}{y + cos ( x )})

Use differentiation rule \frac{\partial}{\partial x} (\frac{f ( x )}{g ( x )}) = \frac{\frac{\partial}{\partial x} ( f ( x )) \times g ( x ) - f ( x ) \times \frac{\partial}{\partial x} ( g ( x ) )}{( g ( x ) ) ^{2}}

f_{x} = \frac{\frac{\partial}{\partial x} ( 2 y ) ( y + cos ( x ) ) - 2 y \times \frac{\partial}{\partial x} ( y + cos ( x ) )}{( y + cos ( x ) ) ^{2}}

Use \frac{\partial}{\partial x} (c) = 0 to find derivative

f_{x} = \frac{0 \times ( y + cos ( x ) ) - 2 y \times \frac{\partial}{\partial x} ( y + cos ( x ) )}{( y + cos ( x ) ) ^{2}}

Evaluate

More Steps

f_{x} = \frac{0 \times ( y + cos ( x ) ) - 2 y ( - sin ( x ) )}{( y + cos ( x ) ) ^{2}}

Any expression multiplied by 0 equals 0

f_{x} = \frac{0 - 2 y ( - sin ( x ) )}{( y + cos ( x ) ) ^{2}}

Evaluate

f_{x} = \frac{0 - ( - 2 y sin ( x ) )}{( y + cos ( x ) ) ^{2}}

Solution

More Steps

f_{x} = \frac{2 y sin ( x )}{( y + cos ( x ) ) ^{2}}