What is the derivative of f(x)=(53)/(0.01x^2+1) ?

The derivative of f(x)=(53)/(0.01x^2+1) is -(1.06x)/((0.01x^2+1)^2)

What is the first derivative of f(x)=(53)/(0.01x^2+1) ?

The first derivative of f(x)=(53)/(0.01x^2+1) is -(1.06x)/((0.01x^2+1)^2)

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derivative of

f (x) = \frac{53}{0.01 x ^{2} + 1}

- \frac{1.06 x}{( 0.01 x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{53}{0.01 x ^{2} + 1})

Take the constant out:

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

= 53 \frac{d}{d x} (\frac{1}{0.01 x ^{2} + 1})

Apply exponent rule:

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

= 53 \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

= 53 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x)

Simplify

53 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x) : - \frac{1.06 x}{( 0.01 x ^{2} + 1 ) ^{2}}

= - \frac{1.06 x}{( 0.01 x ^{2} + 1 ) ^{2}}

\int \frac{e ^{- 4 x}}{x ^{2}} d x

y = - x^{3} + 3, y = x, x = - 1, x = 1

y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (- 3, - 2)

\int \frac{- sin ^{2} ( x )}{cos ( x )} d x

y = sin (sin (x))

What is the derivative of f(x)=(53)/(0.01x^2+1) ?
The derivative of f(x)=(53)/(0.01x^2+1) is -(1.06x)/((0.01x^2+1)^2)
What is the first derivative of f(x)=(53)/(0.01x^2+1) ?
The first derivative of f(x)=(53)/(0.01x^2+1) is -(1.06x)/((0.01x^2+1)^2)