What is the derivative of (500/(1+0.06x^2)) ?

The derivative of (500/(1+0.06x^2)) is -(60x)/((1+0.06x^2)^2)

What is the first derivative of (500/(1+0.06x^2)) ?

The first derivative of (500/(1+0.06x^2)) is -(60x)/((1+0.06x^2)^2)

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\frac{d}{d x} (\frac{500}{1 + 0.06 x ^{2}})

- \frac{60 x}{( 1 + 0.06 x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{500}{1 + 0.06 x ^{2}})

Take the constant out:

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

= 500 \frac{d}{d x} (\frac{1}{1 + 0.06 x ^{2}})

Apply exponent rule:

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

= 500 \frac{d}{d x} ((1 + 0.06 x^{2})^{- 1})

Apply the chain rule:

- \frac{1}{( 1 + 0.06 x ^{2} ) ^{2}} \frac{d}{d x} (1 + 0.06 x^{2})

= - \frac{1}{( 1 + 0.06 x ^{2} ) ^{2}} \frac{d}{d x} (1 + 0.06 x^{2})

\frac{d}{d x} (1 + 0.06 x^{2}) = 0.12 x

= 500 (- \frac{1}{( 1 + 0.06 x ^{2} ) ^{2}} \cdot 0.12 x)

Simplify

500 (- \frac{1}{( 1 + 0.06 x ^{2} ) ^{2}} \cdot 0.12 x) : - \frac{60 x}{( 1 + 0.06 x ^{2} ) ^{2}}

= - \frac{60 x}{( 1 + 0.06 x ^{2} ) ^{2}}

\int \frac{1}{x + 4} d x

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4 x + 4, [- 1, 5.25]

What is the derivative of (500/(1+0.06x^2)) ?
The derivative of (500/(1+0.06x^2)) is -(60x)/((1+0.06x^2)^2)
What is the first derivative of (500/(1+0.06x^2)) ?
The first derivative of (500/(1+0.06x^2)) is -(60x)/((1+0.06x^2)^2)