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Popular >

derivative of f(t)=sqrt(8t^2-t)

Solution

derivative of

f (t) = 8 t^{2} - t

Solution

\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}

Popular Examples

tangent of 4sqrt(x),\at x=4 tangent of

4 x, at x = 4

3x-8ysqrt(x^2+1)(dy)/(dx)=0

3 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

tangent of y=sqrt(x),\at x=1 tangent of

y = x, at x = 1

integral from 7 to e^5 of 1/x

\int_{7}^{e^{5}} \frac{1}{x} d x

(\partial)/(\partial x)(sin(x*y))

\frac{\partial}{\partial x} (sin (x \cdot y))

Frequently Asked Questions (FAQ)

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