What is the derivative of f(x)=(45)/(0.01x^2+1) ?

The derivative of f(x)=(45)/(0.01x^2+1) is -(0.9x)/((0.01x^2+1)^2)

What is the first derivative of f(x)=(45)/(0.01x^2+1) ?

The first derivative of f(x)=(45)/(0.01x^2+1) is -(0.9x)/((0.01x^2+1)^2)

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derivative of

f (x) = \frac{45}{0.01 x ^{2} + 1}

- \frac{0.9 x}{( 0.01 x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{45}{0.01 x ^{2} + 1})

Take the constant out:

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

= 45 \frac{d}{d x} (\frac{1}{0.01 x ^{2} + 1})

Apply exponent rule:

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

= 45 \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

= 45 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x)

Simplify

45 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x) : - \frac{0.9 x}{( 0.01 x ^{2} + 1 ) ^{2}}

= - \frac{0.9 x}{( 0.01 x ^{2} + 1 ) ^{2}}

cos^{2} (5 t)

x^{^{''''}} + 8 x^{^{''}} + 16 x = 0

y = x^{2} - 33, y = 7 - 3 x

25 y^{^{''}} + 9 y = 0

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

What is the derivative of f(x)=(45)/(0.01x^2+1) ?
The derivative of f(x)=(45)/(0.01x^2+1) is -(0.9x)/((0.01x^2+1)^2)
What is the first derivative of f(x)=(45)/(0.01x^2+1) ?
The first derivative of f(x)=(45)/(0.01x^2+1) is -(0.9x)/((0.01x^2+1)^2)