What is the derivative of 2x^2+y^2-ln(xy^2) ?

The derivative of 2x^2+y^2-ln(xy^2) is 4x+2y(dy)/(dx)-(y+2x\frac{dy)/(dx)}{xy}

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

The first derivative of 2x^2+y^2-ln(xy^2) is 4x+2y(dy)/(dx)-(y+2x\frac{dy)/(dx)}{xy}

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

4 x + 2 y \frac{d y}{d x} - \frac{y + 2 x \frac{d y}{d x}}{x y}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (2 x^{2}) = 4 x

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

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

= 4 x + 2 y \frac{d y}{d x} - \frac{y + 2 x \frac{d y}{d x}}{x y}

y = 2 + \frac{1}{x}, (1, 3)

\int \frac{4 x + 1}{2 x ^{2} + x} d x

\frac{\partial}{\partial x} (x^{2} sin (11 t))

\frac{d}{d x} (\frac{x ^{4} - 5 x ^{3} + 8}{x ^{2}})

\frac{d}{d x} (e^{- x^{9}})

What is the derivative of 2x^2+y^2-ln(xy^2) ?
The derivative of 2x^2+y^2-ln(xy^2) is 4x+2y(dy)/(dx)-(y+2x\frac{dy)/(dx)}{xy}
What is the first derivative of 2x^2+y^2-ln(xy^2) ?
The first derivative of 2x^2+y^2-ln(xy^2) is 4x+2y(dy)/(dx)-(y+2x\frac{dy)/(dx)}{xy}