What is the derivative of y=(18)/(x^2+2) ?

The derivative of y=(18)/(x^2+2) is -(36x)/((x^2+2)^2)

What is the first derivative of y=(18)/(x^2+2) ?

The first derivative of y=(18)/(x^2+2) is -(36x)/((x^2+2)^2)

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derivative of

y = \frac{18}{x ^{2} + 2}

- \frac{36 x}{( x ^{2} + 2 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{18}{x ^{2} + 2})

Take the constant out:

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

= 18 \frac{d}{d x} (\frac{1}{x ^{2} + 2})

Apply exponent rule:

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

= 18 \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

= 18 (- \frac{1}{( x ^{2} + 2 ) ^{2}} \cdot 2 x)

Simplify

18 (- \frac{1}{( x ^{2} + 2 ) ^{2}} \cdot 2 x) : - \frac{36 x}{( x ^{2} + 2 ) ^{2}}

= - \frac{36 x}{( x ^{2} + 2 ) ^{2}}

y^{^{'}} = e^{x - y}, y (0) = 2

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\frac{\partial}{\partial x} (ln (x + 9 y + 4 z))

x \to 2 lim (5 x - 7)

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

What is the derivative of y=(18)/(x^2+2) ?
The derivative of y=(18)/(x^2+2) is -(36x)/((x^2+2)^2)
What is the first derivative of y=(18)/(x^2+2) ?
The first derivative of y=(18)/(x^2+2) is -(36x)/((x^2+2)^2)