What is the derivative of y= 1/(t^3+2t^2-1) ?

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

What is the first derivative of y= 1/(t^3+2t^2-1) ?

The first derivative of y= 1/(t^3+2t^2-1) is -(3t^2+4t)/((t^3+2t^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{1}{t ^{3} + 2 t ^{2} - 1}

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

Solution steps

\frac{d}{d t} (\frac{1}{t ^{3} + 2 t ^{2} - 1})

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{1}{( t ^{3} + 2 t ^{2} - 1 ) ^{2}} \frac{d}{d t} (t^{3} + 2 t^{2} - 1)

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

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

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

Simplify

- \frac{1}{( t ^{3} + 2 t ^{2} - 1 ) ^{2}} (3 t^{2} + 4 t) : - \frac{3 t ^{2} + 4 t}{( t ^{3} + 2 t ^{2} - 1 ) ^{2}}

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

x = - 5, x = 1, y = 3 x, y = x^{2} - 4

\frac{d}{d x} (\frac{4 x - 7}{2 x + 1})

\int \frac{6 x ^{2} - 5 x + 3}{x} d x

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

(3, 3), (2, - 5)

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