What is the derivative of 18+(12)/((0.0004x^2+1)^2) ?

The derivative of 18+(12)/((0.0004x^2+1)^2) is -(0.0192x)/((0.0004x^2+1)^3)

What is the first derivative of 18+(12)/((0.0004x^2+1)^2) ?

The first derivative of 18+(12)/((0.0004x^2+1)^2) is -(0.0192x)/((0.0004x^2+1)^3)

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derivative of

18 + \frac{12}{( 0.0004 x ^{2} + 1 ) ^{2}}

- \frac{0.0192 x}{( 0.0004 x ^{2} + 1 ) ^{3}}

Solution steps

\frac{d}{d x} (18 + \frac{12}{( 0.0004 x ^{2} + 1 ) ^{2}})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (18) + \frac{d}{d x} (\frac{12}{( 0.0004 x ^{2} + 1 ) ^{2}})

\frac{d}{d x} (18) = 0

\frac{d}{d x} (\frac{12}{( 0.0004 x ^{2} + 1 ) ^{2}}) = - \frac{0.0192 x}{( 0.0004 x ^{2} + 1 ) ^{3}}

= 0 - \frac{0.0192 x}{( 0.0004 x ^{2} + 1 ) ^{3}}

Simplify

= - \frac{0.0192 x}{( 0.0004 x ^{2} + 1 ) ^{3}}

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\frac{\partial}{\partial y} (2 x^{4 y})

\frac{d}{d x} (9 tan (x) - 9 cot (x))

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x \to 0 lim (2 x^{2} + 4 x - 2)

What is the derivative of 18+(12)/((0.0004x^2+1)^2) ?
The derivative of 18+(12)/((0.0004x^2+1)^2) is -(0.0192x)/((0.0004x^2+1)^3)
What is the first derivative of 18+(12)/((0.0004x^2+1)^2) ?
The first derivative of 18+(12)/((0.0004x^2+1)^2) is -(0.0192x)/((0.0004x^2+1)^3)