What is the derivative of f(x)=(216)/(x^2+36) ?

The derivative of f(x)=(216)/(x^2+36) is -(432x)/((x^2+36)^2)

What is the first derivative of f(x)=(216)/(x^2+36) ?

The first derivative of f(x)=(216)/(x^2+36) is -(432x)/((x^2+36)^2)

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derivative of

f (x) = \frac{216}{x ^{2} + 36}

- \frac{432 x}{( x ^{2} + 36 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{216}{x ^{2} + 36})

Take the constant out:

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

= 216 \frac{d}{d x} (\frac{1}{x ^{2} + 36})

Apply exponent rule:

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

= 216 \frac{d}{d x} ((x^{2} + 36)^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{2} + 36 ) ^{2}} \frac{d}{d x} (x^{2} + 36)

= - \frac{1}{( x ^{2} + 36 ) ^{2}} \frac{d}{d x} (x^{2} + 36)

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

= 216 (- \frac{1}{( x ^{2} + 36 ) ^{2}} \cdot 2 x)

Simplify

216 (- \frac{1}{( x ^{2} + 36 ) ^{2}} \cdot 2 x) : - \frac{432 x}{( x ^{2} + 36 ) ^{2}}

= - \frac{432 x}{( x ^{2} + 36 ) ^{2}}

\frac{d x}{d t} = \frac{4}{x e ^{t + 3 x}}

- \frac{2 ( z ^{2} + 1 )}{z ^{2}}

\frac{d}{d t} (a t^{5} - 5 b t^{3})

\frac{\partial}{\partial x} (5 x y^{3})

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

What is the derivative of f(x)=(216)/(x^2+36) ?
The derivative of f(x)=(216)/(x^2+36) is -(432x)/((x^2+36)^2)
What is the first derivative of f(x)=(216)/(x^2+36) ?
The first derivative of f(x)=(216)/(x^2+36) is -(432x)/((x^2+36)^2)