What is the derivative of 10+(50t)/(2t^2+9) ?

The derivative of 10+(50t)/(2t^2+9) is (50(-2t^2+9))/((2t^2+9)^2)

What is the first derivative of 10+(50t)/(2t^2+9) ?

The first derivative of 10+(50t)/(2t^2+9) is (50(-2t^2+9))/((2t^2+9)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

derivative of

10 + \frac{50 t}{2 t ^{2} + 9}

\frac{50 ( - 2 t ^{2} + 9 )}{( 2 t ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d t} (10 + \frac{50 t}{2 t ^{2} + 9})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (10) + \frac{d}{d t} (\frac{50 t}{2 t ^{2} + 9})

\frac{d}{d t} (10) = 0

\frac{d}{d t} (\frac{50 t}{2 t ^{2} + 9}) = \frac{50 ( - 2 t ^{2} + 9 )}{( 2 t ^{2} + 9 ) ^{2}}

= 0 + \frac{50 ( - 2 t ^{2} + 9 )}{( 2 t ^{2} + 9 ) ^{2}}

Simplify

= \frac{50 ( - 2 t ^{2} + 9 )}{( 2 t ^{2} + 9 ) ^{2}}

What is the derivative of 10+(50t)/(2t^2+9) ?
The derivative of 10+(50t)/(2t^2+9) is (50(-2t^2+9))/((2t^2+9)^2)
What is the first derivative of 10+(50t)/(2t^2+9) ?
The first derivative of 10+(50t)/(2t^2+9) is (50(-2t^2+9))/((2t^2+9)^2)