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derivative of (2t^2+1)/t

Solution

derivative of

\frac{2 t ^{2} + 1}{t}

Solution

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{d t}{d t} = 1

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

Simplify

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

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

Popular Examples

derivative of xe^{-sqrt(x)}

\frac{d}{d x} (x e^{- x})

taylor f(x)=cos(4x)taylor

f (x) = cos (4 x)

tangent of f(x)=2sqrt(x),\at x=4 tangent of

f (x) = 2 x, at x = 4

y^'=(3x^2+2x-4)/(2y-4),y(2)=0

y^{^{'}} = \frac{3 x ^{2} + 2 x - 4}{2 y - 4}, y (2) = 0

inverse oflaplace (2s)/((s^2+1)^2(s+1))inverse laplace

\frac{2 s}{( s ^{2} + 1 ) ^{2} ( s + 1 )}

Frequently Asked Questions (FAQ)

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