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

The derivative of f(x)=(6x)/(sqrt(6x^2+5)) is (30)/((6x^2+5)sqrt(6x^2+5))

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

The first derivative of f(x)=(6x)/(sqrt(6x^2+5)) is (30)/((6x^2+5)sqrt(6x^2+5))

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derivative of

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

\frac{30}{( 6 x ^{2} + 5 ) 6 x ^{2} + 5}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

= 6 \cdot \frac{1 \cdot 6 x ^{2} + 5 - \frac{6 x}{6 x ^{2} + 5} x}{( 6 x ^{2} + 5 ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot 6 x ^{2} + 5 - \frac{6 x}{6 x ^{2} + 5} x}{( 6 x ^{2} + 5 ) ^{2}} : \frac{30}{( 6 x ^{2} + 5 ) 6 x ^{2} + 5}

= \frac{30}{( 6 x ^{2} + 5 ) 6 x ^{2} + 5}

\frac{\partial}{\partial x} (sin (π (6 x - 5 y)))

x + \frac{sin ( y )}{cos ( x )} y^{^{'}} = 0

\frac{e ^{x} + e ^{- x}}{6}

f (x) = \frac{3}{( x ^{2} - 5 x ) ^{2}}

x \to 0 lim (\frac{x ^{\frac{2}{3}}}{x})

What is the derivative of f(x)=(6x)/(sqrt(6x^2+5)) ?
The derivative of f(x)=(6x)/(sqrt(6x^2+5)) is (30)/((6x^2+5)sqrt(6x^2+5))
What is the first derivative of f(x)=(6x)/(sqrt(6x^2+5)) ?
The first derivative of f(x)=(6x)/(sqrt(6x^2+5)) is (30)/((6x^2+5)sqrt(6x^2+5))