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

The derivative of 1/(t^2+5t-4) is -(2t+5)/((t^2+5t-4)^2)

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

The first derivative of 1/(t^2+5t-4) is -(2t+5)/((t^2+5t-4)^2)

Popular >

derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (3 x^{\frac{5}{3}} + 4 x^{- \frac{2}{3}} - 7 x^{\frac{3}{4}})

\int_{0}^{\infty} \frac{6}{x ^{2} + 4} d x

\int \frac{1}{x ^{2} 9 x ^{2} + 4} d x

\frac{d}{d x} ((6 x^{2} - 10 x) e^{x})

n = 1 \sum \infty e^{(2 n - 1)}

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