What is the derivative of f(0)=(cos(x))/(x^2-1) ?

The derivative of f(0)=(cos(x))/(x^2-1) is (-sin(x)(x^2-1)-2xcos(x))/((x^2-1)^2)

What is the first derivative of f(0)=(cos(x))/(x^2-1) ?

The first derivative of f(0)=(cos(x))/(x^2-1) is (-sin(x)(x^2-1)-2xcos(x))/((x^2-1)^2)

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derivative of

f (0) = \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}}

\int_{1}^{4} 7 e^{x} d x

n = 1 \sum \infty \frac{2 ^{n}}{1 0 ^{n}}

\frac{\partial ^{2}}{\partial y ^{2}} (x y^{10} + x^{2} + y^{4})

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

\frac{1}{s ^{3} + s ^{2} + s}

What is the derivative of f(0)=(cos(x))/(x^2-1) ?
The derivative of f(0)=(cos(x))/(x^2-1) is (-sin(x)(x^2-1)-2xcos(x))/((x^2-1)^2)
What is the first derivative of f(0)=(cos(x))/(x^2-1) ?
The first derivative of f(0)=(cos(x))/(x^2-1) is (-sin(x)(x^2-1)-2xcos(x))/((x^2-1)^2)