What is the derivative of (500)/(5t^2+10t+100) ?

The derivative of (500)/(5t^2+10t+100) is -(200(t+1))/((t^2+2t+20)^2)

What is the first derivative of (500)/(5t^2+10t+100) ?

The first derivative of (500)/(5t^2+10t+100) is -(200(t+1))/((t^2+2t+20)^2)

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derivative of

\frac{500}{5 t ^{2} + 10 t + 100}

- \frac{200 ( t + 1 )}{( t ^{2} + 2 t + 20 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{500}{5 t ^{2} + 10 t + 100})

Take the constant out:

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

= 500 \frac{d}{d t} (\frac{1}{5 t ^{2} + 10 t + 100})

Apply exponent rule:

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

= 500 \frac{d}{d t} ((5 t^{2} + 10 t + 100)^{- 1})

Apply the chain rule:

- \frac{1}{( 5 t ^{2} + 10 t + 100 ) ^{2}} \frac{d}{d t} (5 t^{2} + 10 t + 100)

= - \frac{1}{( 5 t ^{2} + 10 t + 100 ) ^{2}} \frac{d}{d t} (5 t^{2} + 10 t + 100)

\frac{d}{d t} (5 t^{2} + 10 t + 100) = 10 t + 10

= 500 (- \frac{1}{( 5 t ^{2} + 10 t + 100 ) ^{2}} (10 t + 10))

Simplify

500 (- \frac{1}{( 5 t ^{2} + 10 t + 100 ) ^{2}} (10 t + 10)) : - \frac{200 ( t + 1 )}{( t ^{2} + 2 t + 20 ) ^{2}}

= - \frac{200 ( t + 1 )}{( t ^{2} + 2 t + 20 ) ^{2}}

\int \frac{\frac{1}{x}}{ln ( x ) ln ^{2} ( x ) - 1} d x

\frac{d y}{d x} - 7 y = e^{x}

\int \frac{x}{x ^{2} - 2 x + 2} d x

x \to 5 + lim (\frac{3 x}{4} + 1)

\frac{d}{d x} (ln (x - a))

What is the derivative of (500)/(5t^2+10t+100) ?
The derivative of (500)/(5t^2+10t+100) is -(200(t+1))/((t^2+2t+20)^2)
What is the first derivative of (500)/(5t^2+10t+100) ?
The first derivative of (500)/(5t^2+10t+100) is -(200(t+1))/((t^2+2t+20)^2)