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Popular >

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

Solution

derivative of

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

Solution

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{4 t - 1}{2 2 t ^{2} - t}

Popular Examples

y^'+1/(x+1)y=x^{-2}

y^{^{'}} + \frac{1}{x + 1} y = x^{- 2}

tangent of f(x)=(3x-2)(x+4),(1,5)tangent of

f (x) = (3 x - 2) (x + 4), (1, 5)

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

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x \to \infty lim (\frac{- x}{x})

integral of ((3x-5))/(x^2+4x)

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

Frequently Asked Questions (FAQ)

What is the derivative of f(t)=sqrt(2t^2-t) ?
The derivative of f(t)=sqrt(2t^2-t) is (4t-1)/(2sqrt(2t^2-t))
What is the first derivative of f(t)=sqrt(2t^2-t) ?
The first derivative of f(t)=sqrt(2t^2-t) is (4t-1)/(2sqrt(2t^2-t))