What is the derivative of-(49x/(36y)) ?

The derivative of-(49x/(36y)) is -(49(y-x\frac{dy)/(dx))}{36y^2}

What is the first derivative of-(49x/(36y)) ?

The first derivative of-(49x/(36y)) is -(49(y-x\frac{dy)/(dx))}{36y^2}

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\frac{d}{d x} (- \frac{49 x}{36 y})

- \frac{49 ( y - x \frac{d y}{d x} )}{36 y ^{2}}

Solution steps

\frac{d}{d x} (- \frac{49 x}{36 y})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - \frac{49}{36} \frac{d}{d x} (\frac{x}{y})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= - \frac{49}{36} \cdot \frac{\frac{d x}{d x} y - \frac{d y}{d x} x}{y ^{2}}

\frac{d x}{d x} = 1

= - \frac{49}{36} \cdot \frac{1 \cdot y - \frac{d y}{d x} x}{y ^{2}}

Simplify

- \frac{49}{36} \cdot \frac{1 \cdot y - \frac{d y}{d x} x}{y ^{2}} : - \frac{49 ( y - x \frac{d y}{d x} )}{36 y ^{2}}

= - \frac{49 ( y - x \frac{d y}{d x} )}{36 y ^{2}}

7 x + 4, at x = 3

\frac{d}{d x} (- e^{5 x})

\frac{\partial}{\partial y} (x^{2} + y^{2} + x^{2} y + 4)

\int \frac{x ^{2}}{x ^{6} + 1} d x

f (x) = x^{3} - x

What is the derivative of-(49x/(36y)) ?
The derivative of-(49x/(36y)) is -(49(y-x\frac{dy)/(dx))}{36y^2}
What is the first derivative of-(49x/(36y)) ?
The first derivative of-(49x/(36y)) is -(49(y-x\frac{dy)/(dx))}{36y^2}