What is the derivative of f(x)=(cos(x))/(x^2-1) ?

The derivative of f(x)=(cos(x))/(x^2-1) is (-sin(x)(x^2-1)-2xcos(x))/((x^2-1)^2)

What is the first derivative of f(x)=(cos(x))/(x^2-1) ?

The first derivative of f(x)=(cos(x))/(x^2-1) is (-sin(x)(x^2-1)-2xcos(x))/((x^2-1)^2)

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derivative of

f (x) = \frac{cos ( x )}{x ^{2} - 1}

\frac{- sin ( x ) ( x ^{2} - 1 ) - 2 x cos ( x )}{( x ^{2} - 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{cos ( x )}{x ^{2} - 1})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (cos (x)) = - sin (x)

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

= \frac{( - sin ( x ) ) ( x ^{2} - 1 ) - 2 x cos ( x )}{( x ^{2} - 1 ) ^{2}}

Simplify

= \frac{- sin ( x ) ( x ^{2} - 1 ) - 2 x cos ( x )}{( x ^{2} - 1 ) ^{2}}

f (x) = \frac{( 4 x ^{2} - \frac{1}{2} x )}{( 9 x + 8 )}

f (x) = sin (π - x), at x = \frac{π}{6}

\frac{\partial}{\partial y} (x^{2} + y^{2} + 1)

\frac{d}{d x} (\frac{2 x}{x + 1})

x - 6

What is the derivative of f(x)=(cos(x))/(x^2-1) ?
The derivative of f(x)=(cos(x))/(x^2-1) is (-sin(x)(x^2-1)-2xcos(x))/((x^2-1)^2)
What is the first derivative of f(x)=(cos(x))/(x^2-1) ?
The first derivative of f(x)=(cos(x))/(x^2-1) is (-sin(x)(x^2-1)-2xcos(x))/((x^2-1)^2)