What is the derivative of y=(t^2+4)/(t^4-5t^2+1) ?

The derivative of y=(t^2+4)/(t^4-5t^2+1) is (-2t^5-16t^3+42t)/((t^4-5t^2+1)^2)

What is the first derivative of y=(t^2+4)/(t^4-5t^2+1) ?

The first derivative of y=(t^2+4)/(t^4-5t^2+1) is (-2t^5-16t^3+42t)/((t^4-5t^2+1)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

y = \frac{t ^{2} + 4}{t ^{4} - 5 t ^{2} + 1}

\frac{- 2 t ^{5} - 16 t ^{3} + 42 t}{( t ^{4} - 5 t ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{t ^{2} + 4}{t ^{4} - 5 t ^{2} + 1})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d t} ( t ^{2} + 4 ) ( t ^{4} - 5 t ^{2} + 1 ) - \frac{d}{d t} ( t ^{4} - 5 t ^{2} + 1 ) ( t ^{2} + 4 )}{( t ^{4} - 5 t ^{2} + 1 ) ^{2}}

\frac{d}{d t} (t^{2} + 4) = 2 t

\frac{d}{d t} (t^{4} - 5 t^{2} + 1) = 4 t^{3} - 10 t

= \frac{2 t ( t ^{4} - 5 t ^{2} + 1 ) - ( 4 t ^{3} - 10 t ) ( t ^{2} + 4 )}{( t ^{4} - 5 t ^{2} + 1 ) ^{2}}

Expand

2 t (t^{4} - 5 t^{2} + 1) - (4 t^{3} - 10 t) (t^{2} + 4) : - 2 t^{5} - 16 t^{3} + 42 t

= \frac{- 2 t ^{5} - 16 t ^{3} + 42 t}{( t ^{4} - 5 t ^{2} + 1 ) ^{2}}

\int 2 x (x^{2} + 7)^{4} d x

(y + x^{2} + y^{2}) d x - x d y = 0

f (x) = e^{x} - 4 x^{3}

x \to - 6 lim (2 x + 1)

e^{2 x}

What is the derivative of y=(t^2+4)/(t^4-5t^2+1) ?
The derivative of y=(t^2+4)/(t^4-5t^2+1) is (-2t^5-16t^3+42t)/((t^4-5t^2+1)^2)
What is the first derivative of y=(t^2+4)/(t^4-5t^2+1) ?
The first derivative of y=(t^2+4)/(t^4-5t^2+1) is (-2t^5-16t^3+42t)/((t^4-5t^2+1)^2)