What is the derivative of s(t)= 1/(t^2+6t-7) ?

The derivative of s(t)= 1/(t^2+6t-7) is -(2t+6)/((t^2+6t-7)^2)

What is the first derivative of s(t)= 1/(t^2+6t-7) ?

The first derivative of s(t)= 1/(t^2+6t-7) is -(2t+6)/((t^2+6t-7)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

s (t) = \frac{1}{t ^{2} + 6 t - 7}

- \frac{2 t + 6}{( t ^{2} + 6 t - 7 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{1}{t ^{2} + 6 t - 7})

Apply exponent rule:

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

= \frac{d}{d t} ((t^{2} + 6 t - 7)^{- 1})

Apply the chain rule:

- \frac{1}{( t ^{2} + 6 t - 7 ) ^{2}} \frac{d}{d t} (t^{2} + 6 t - 7)

= - \frac{1}{( t ^{2} + 6 t - 7 ) ^{2}} \frac{d}{d t} (t^{2} + 6 t - 7)

\frac{d}{d t} (t^{2} + 6 t - 7) = 2 t + 6

= - \frac{1}{( t ^{2} + 6 t - 7 ) ^{2}} (2 t + 6)

Simplify

- \frac{1}{( t ^{2} + 6 t - 7 ) ^{2}} (2 t + 6) : - \frac{2 t + 6}{( t ^{2} + 6 t - 7 ) ^{2}}

= - \frac{2 t + 6}{( t ^{2} + 6 t - 7 ) ^{2}}

\int \frac{ln ( 4 x )}{x} d x

\frac{\partial}{\partial x} (- \frac{2 y}{x ^{2} + y ^{2} + 1})

\int e^{7 x} cos (8 x) d x

\int_{0}^{2 π} cos (x) d x

\frac{\partial}{\partial x} ((4 x - 5 y)^{\frac{3}{2}})

What is the derivative of s(t)= 1/(t^2+6t-7) ?
The derivative of s(t)= 1/(t^2+6t-7) is -(2t+6)/((t^2+6t-7)^2)
What is the first derivative of s(t)= 1/(t^2+6t-7) ?
The first derivative of s(t)= 1/(t^2+6t-7) is -(2t+6)/((t^2+6t-7)^2)