What is the derivative of (x^2/(x^2-49)) ?

The derivative of (x^2/(x^2-49)) is -(98x)/((x^2-49)^2)

What is the first derivative of (x^2/(x^2-49)) ?

The first derivative of (x^2/(x^2-49)) is -(98x)/((x^2-49)^2)

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

- \frac{98 x}{( x ^{2} - 49 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 x ( x ^{2} - 49 ) - 2 x x ^{2}}{( x ^{2} - 49 ) ^{2}} : - \frac{98 x}{( x ^{2} - 49 ) ^{2}}

= - \frac{98 x}{( x ^{2} - 49 ) ^{2}}

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

\int \frac{e ^{x}}{a + b e ^{x}} d x

(3 + x^{2}) y^{^{'}} - 2 x y = 0

\int_{0}^{1} x x^{2} + 1 d x

\int (2 x^{3} - x^{2} + 3 x - 7) d x

What is the derivative of (x^2/(x^2-49)) ?
The derivative of (x^2/(x^2-49)) is -(98x)/((x^2-49)^2)
What is the first derivative of (x^2/(x^2-49)) ?
The first derivative of (x^2/(x^2-49)) is -(98x)/((x^2-49)^2)