What is the derivative of 5/(x^2-2) ?

The derivative of 5/(x^2-2) is -(10x)/((x^2-2)^2)

What is the first derivative of 5/(x^2-2) ?

The first derivative of 5/(x^2-2) is -(10x)/((x^2-2)^2)

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\frac{d}{d x} (\frac{5}{x ^{2} - 2})

- \frac{10 x}{( x ^{2} - 2 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{5}{x ^{2} - 2})

Take the constant out:

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

= 5 \frac{d}{d x} (\frac{1}{x ^{2} - 2})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 5 \frac{d}{d x} ((x^{2} - 2)^{- 1})

Apply the chain rule:

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

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

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

= 5 (- \frac{1}{( x ^{2} - 2 ) ^{2}} \cdot 2 x)

Simplify

5 (- \frac{1}{( x ^{2} - 2 ) ^{2}} \cdot 2 x) : - \frac{10 x}{( x ^{2} - 2 ) ^{2}}

= - \frac{10 x}{( x ^{2} - 2 ) ^{2}}

\frac{\partial}{\partial x} (x^{3} + e^{y^{2}})

y^{^{'}} + y = sin (x)

n = 1 \sum \infty n

d er i v a t i v e f (x) = ln (4 x)

y^{^{''}} + y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 0

What is the derivative of 5/(x^2-2) ?
The derivative of 5/(x^2-2) is -(10x)/((x^2-2)^2)
What is the first derivative of 5/(x^2-2) ?
The first derivative of 5/(x^2-2) is -(10x)/((x^2-2)^2)