What is the derivative of f(x)=(60)/(0.01x^2+1) ?

The derivative of f(x)=(60)/(0.01x^2+1) is -(1.2x)/((0.01x^2+1)^2)

What is the first derivative of f(x)=(60)/(0.01x^2+1) ?

The first derivative of f(x)=(60)/(0.01x^2+1) is -(1.2x)/((0.01x^2+1)^2)

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derivative of

f (x) = \frac{60}{0.01 x ^{2} + 1}

- \frac{1.2 x}{( 0.01 x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{60}{0.01 x ^{2} + 1})

Take the constant out:

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

= 60 \frac{d}{d x} (\frac{1}{0.01 x ^{2} + 1})

Apply exponent rule:

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

= 60 \frac{d}{d x} ((0.01 x^{2} + 1)^{- 1})

Apply the chain rule:

- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \frac{d}{d x} (0.01 x^{2} + 1)

= - \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \frac{d}{d x} (0.01 x^{2} + 1)

\frac{d}{d x} (0.01 x^{2} + 1) = 0.02 x

= 60 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x)

Simplify

60 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x) : - \frac{1.2 x}{( 0.01 x ^{2} + 1 ) ^{2}}

= - \frac{1.2 x}{( 0.01 x ^{2} + 1 ) ^{2}}

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

n = 1 \sum \infty \frac{50}{5 ^{n}}

e^{l n (s e c (2 x))}

\frac{d}{d x} (\frac{sin ( 2 x )}{2})

y^{^{''}} - 5 y^{^{'}} + 6 y = x e^{x}

What is the derivative of f(x)=(60)/(0.01x^2+1) ?
The derivative of f(x)=(60)/(0.01x^2+1) is -(1.2x)/((0.01x^2+1)^2)
What is the first derivative of f(x)=(60)/(0.01x^2+1) ?
The first derivative of f(x)=(60)/(0.01x^2+1) is -(1.2x)/((0.01x^2+1)^2)