What is the derivative of 1/(5t^2-6t+2) ?

The derivative of 1/(5t^2-6t+2) is -(10t-6)/((5t^2-6t+2)^2)

What is the first derivative of 1/(5t^2-6t+2) ?

The first derivative of 1/(5t^2-6t+2) is -(10t-6)/((5t^2-6t+2)^2)

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derivative of

\frac{1}{5 t ^{2} - 6 t + 2}

- \frac{10 t - 6}{( 5 t ^{2} - 6 t + 2 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{1}{5 t ^{2} - 6 t + 2})

Apply exponent rule:

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

= \frac{d}{d t} ((5 t^{2} - 6 t + 2)^{- 1})

Apply the chain rule:

- \frac{1}{( 5 t ^{2} - 6 t + 2 ) ^{2}} \frac{d}{d t} (5 t^{2} - 6 t + 2)

= - \frac{1}{( 5 t ^{2} - 6 t + 2 ) ^{2}} \frac{d}{d t} (5 t^{2} - 6 t + 2)

\frac{d}{d t} (5 t^{2} - 6 t + 2) = 10 t - 6

= - \frac{1}{( 5 t ^{2} - 6 t + 2 ) ^{2}} (10 t - 6)

Simplify

- \frac{1}{( 5 t ^{2} - 6 t + 2 ) ^{2}} (10 t - 6) : - \frac{10 t - 6}{( 5 t ^{2} - 6 t + 2 ) ^{2}}

= - \frac{10 t - 6}{( 5 t ^{2} - 6 t + 2 ) ^{2}}

\int \frac{y}{y + 1} d y

\int \frac{tan ^{3} ( ln ( x ) )}{5 x} d x

y^{'} = \frac{x y + y ^{2}}{x ^{2}}

\int \frac{e ^{3 x}}{4 - 9 e ^{6 x}} d x

What is the derivative of 1/(5t^2-6t+2) ?
The derivative of 1/(5t^2-6t+2) is -(10t-6)/((5t^2-6t+2)^2)
What is the first derivative of 1/(5t^2-6t+2) ?
The first derivative of 1/(5t^2-6t+2) is -(10t-6)/((5t^2-6t+2)^2)