What is the derivative of 500(3+(5t)/(30+t^2)) ?

The derivative of 500(3+(5t)/(30+t^2)) is (2500(-t^2+30))/((t^2+30)^2)

What is the first derivative of 500(3+(5t)/(30+t^2)) ?

The first derivative of 500(3+(5t)/(30+t^2)) is (2500(-t^2+30))/((t^2+30)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

500 (3 + \frac{5 t}{30 + t ^{2}})

\frac{2500 ( - t ^{2} + 30 )}{( t ^{2} + 30 ) ^{2}}

Solution steps

\frac{d}{d t} (500 (3 + \frac{5 t}{30 + t ^{2}}))

Take the constant out:

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

= 500 \frac{d}{d t} (3 + \frac{5 t}{30 + t ^{2}})

Apply the Sum/Difference Rule:

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

= 500 (\frac{d}{d t} (3) + \frac{d}{d t} (\frac{5 t}{30 + t ^{2}}))

\frac{d}{d t} (3) = 0

\frac{d}{d t} (\frac{5 t}{30 + t ^{2}}) = \frac{5 ( 30 - t ^{2} )}{( 30 + t ^{2} ) ^{2}}

= 500 (0 + \frac{5 ( 30 - t ^{2} )}{( 30 + t ^{2} ) ^{2}})

Simplify

500 (0 + \frac{5 ( 30 - t ^{2} )}{( 30 + t ^{2} ) ^{2}}) : \frac{2500 ( - t ^{2} + 30 )}{( t ^{2} + 30 ) ^{2}}

= \frac{2500 ( - t ^{2} + 30 )}{( t ^{2} + 30 ) ^{2}}

\int π (y + 2)^{2} d y

\frac{\partial}{\partial y} (x^{3} + x^{2} y^{3} - 2 y^{2})

\frac{d ^{2}}{d x ^{2}} (e^{6 x})

\int \frac{x ^{2}}{e ^{x^{3}}} d x

\frac{d}{d x} (ln (16))

What is the derivative of 500(3+(5t)/(30+t^2)) ?
The derivative of 500(3+(5t)/(30+t^2)) is (2500(-t^2+30))/((t^2+30)^2)
What is the first derivative of 500(3+(5t)/(30+t^2)) ?
The first derivative of 500(3+(5t)/(30+t^2)) is (2500(-t^2+30))/((t^2+30)^2)