What is the derivative of f(t)=(5t^2+t+6)/(sqrt(t)) ?

The derivative of f(t)=(5t^2+t+6)/(sqrt(t)) is (15t^2+t-6)/(2tsqrt(t))

What is the first derivative of f(t)=(5t^2+t+6)/(sqrt(t)) ?

The first derivative of f(t)=(5t^2+t+6)/(sqrt(t)) is (15t^2+t-6)/(2tsqrt(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

derivative of

f (t) = \frac{5 t ^{2} + t + 6}{t}

\frac{15 t ^{2} + t - 6}{2 t t}

Solution steps

\frac{d}{d t} (\frac{5 t ^{2} + t + 6}{t})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 5 t ^{2} + t + 6 ) t - \frac{d}{d t} ( t ) ( 5 t ^{2} + t + 6 )}{( t ) ^{2}}

\frac{d}{d t} (5 t^{2} + t + 6) = 10 t + 1

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

= \frac{( 10 t + 1 ) t - \frac{1}{2 t} ( 5 t ^{2} + t + 6 )}{( t ) ^{2}}

Simplify

\frac{( 10 t + 1 ) t - \frac{1}{2 t} ( 5 t ^{2} + t + 6 )}{( t ) ^{2}} : \frac{15 t ^{2} + t - 6}{2 t t}

= \frac{15 t ^{2} + t - 6}{2 t t}

\int \frac{7 x}{( 1 - x ^{2} ) ^{3}} d x

x \to π lim (\frac{cos ( x )}{x})

y^{^{'}} = 2 x y + 3 y - 4 x - 6

x \to \infty lim (- e^{x})

\int e^{2 x} 1 - 4 e^{2 x} d x

What is the derivative of f(t)=(5t^2+t+6)/(sqrt(t)) ?
The derivative of f(t)=(5t^2+t+6)/(sqrt(t)) is (15t^2+t-6)/(2tsqrt(t))
What is the first derivative of f(t)=(5t^2+t+6)/(sqrt(t)) ?
The first derivative of f(t)=(5t^2+t+6)/(sqrt(t)) is (15t^2+t-6)/(2tsqrt(t))