What is the derivative of-1/(1+y^2-cos(x)+2xy) ?

The derivative of-1/(1+y^2-cos(x)+2xy) is (2y\frac{dy)/(dx)}{(1+y^2)^2}+sin(x)+2(y+x(dy)/(dx))

What is the first derivative of-1/(1+y^2-cos(x)+2xy) ?

The first derivative of-1/(1+y^2-cos(x)+2xy) is (2y\frac{dy)/(dx)}{(1+y^2)^2}+sin(x)+2(y+x(dy)/(dx))

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\frac{d}{d x} (- \frac{1}{1 + y ^{2}} - cos (x) + 2 x y)

\frac{2 y \frac{d y}{d x}}{( 1 + y ^{2} ) ^{2}} + sin (x) + 2 (y + x \frac{d y}{d x})

Solution steps

\frac{d}{d x} (- \frac{1}{1 + y ^{2}} - cos (x) + 2 x y)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d x} (\frac{1}{1 + y ^{2}}) - \frac{d}{d x} (cos (x)) + \frac{d}{d x} (2 x y)

\frac{d}{d x} (\frac{1}{1 + y ^{2}}) = - \frac{2 y \frac{d y}{d x}}{( 1 + y ^{2} ) ^{2}}

\frac{d}{d x} (cos (x)) = - sin (x)

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

= - (- \frac{2 y \frac{d y}{d x}}{( 1 + y ^{2} ) ^{2}}) - (- sin (x)) + 2 (y + x \frac{d y}{d x})

Simplify

= \frac{2 y \frac{d y}{d x}}{( 1 + y ^{2} ) ^{2}} + sin (x) + 2 (y + x \frac{d y}{d x})

\frac{x}{9 + x}

x \to 1 lim (\frac{x - 1}{1 - x ^{2}})

\int x^{4} sin (2 x) d x

f (x) = arcsin (6 x^{4})

n = 1 \sum \infty n e^{n}

What is the derivative of-1/(1+y^2-cos(x)+2xy) ?
The derivative of-1/(1+y^2-cos(x)+2xy) is (2y\frac{dy)/(dx)}{(1+y^2)^2}+sin(x)+2(y+x(dy)/(dx))
What is the first derivative of-1/(1+y^2-cos(x)+2xy) ?
The first derivative of-1/(1+y^2-cos(x)+2xy) is (2y\frac{dy)/(dx)}{(1+y^2)^2}+sin(x)+2(y+x(dy)/(dx))