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

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

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

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

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derivative of

\frac{1}{t ^{2} + 6 t - 2}

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (x^{4} + 3 x^{2} - 6)

x \to 0 lim (\frac{5 x}{x ^{2} + 5 x})

\frac{d y}{d x} + 2 y = e^{x} y^{- 6}

\int \frac{( ln ( x ) ) 2}{x} d x

\int_{0}^{\frac{π}{2}} e^{s i n (5 π x)} cos (5 π x) d x

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