What is the derivative of ln(8-t^2) ?

The derivative of ln(8-t^2) is -(2t)/(8-t^2)

What is the first derivative of ln(8-t^2) ?

The first derivative of ln(8-t^2) is -(2t)/(8-t^2)

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derivative of

ln (8 - t^{2})

- \frac{2 t}{8 - t ^{2}}

Solution steps

\frac{d}{d t} (ln (8 - t^{2}))

Apply the chain rule:

\frac{1}{8 - t ^{2}} \frac{d}{d t} (8 - t^{2})

= \frac{1}{8 - t ^{2}} \frac{d}{d t} (8 - t^{2})

\frac{d}{d t} (8 - t^{2}) = - 2 t

= \frac{1}{8 - t ^{2}} (- 2 t)

Simplify

\frac{1}{8 - t ^{2}} (- 2 t) : - \frac{2 t}{8 - t ^{2}}

= - \frac{2 t}{8 - t ^{2}}

\int cot^{6} (x) csc^{2} (x) d x

\int \frac{1}{x ^{0.5}} d x

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

(- 5) (0.1)

x y^{2} y^{^{'}} = y^{3} - x^{3}, y (1) = 3

What is the derivative of ln(8-t^2) ?
The derivative of ln(8-t^2) is -(2t)/(8-t^2)
What is the first derivative of ln(8-t^2) ?
The first derivative of ln(8-t^2) is -(2t)/(8-t^2)