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

The derivative of sqrt(8t^2-t) is (16t-1)/(2sqrt(8t^2-t))

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

The first derivative of sqrt(8t^2-t) is (16t-1)/(2sqrt(8t^2-t))

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derivative of

8 t^{2} - t

\frac{16 t - 1}{2 8 t ^{2} - t}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d t} (8 t^{2} - t) = 16 t - 1

= \frac{1}{2 8 t ^{2} - t} (16 t - 1)

Simplify

\frac{1}{2 8 t ^{2} - t} (16 t - 1) : \frac{16 t - 1}{2 8 t ^{2} - t}

= \frac{16 t - 1}{2 8 t ^{2} - t}

\frac{d}{d x} (5 e^{x^{3}})

(3 x + 5)^{2}

(2 x - 1) d x + (7 y + 8) d y = 0

x \to - 5 lim (4)

\int \frac{( 1 + x ) ^{9}}{x} d x

What is the derivative of sqrt(8t^2-t) ?
The derivative of sqrt(8t^2-t) is (16t-1)/(2sqrt(8t^2-t))
What is the first derivative of sqrt(8t^2-t) ?
The first derivative of sqrt(8t^2-t) is (16t-1)/(2sqrt(8t^2-t))