What is the derivative of (-cos(x))/(4x^2+2x-3) ?

The derivative of (-cos(x))/(4x^2+2x-3) is -(-sin(x)(4x^2+2x-3)-cos(x)(8x+2))/((4x^2+2x-3)^2)

What is the first derivative of (-cos(x))/(4x^2+2x-3) ?

The first derivative of (-cos(x))/(4x^2+2x-3) is -(-sin(x)(4x^2+2x-3)-cos(x)(8x+2))/((4x^2+2x-3)^2)

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derivative of

\frac{- cos ( x )}{4 x ^{2} + 2 x - 3}

- \frac{- sin ( x ) ( 4 x ^{2} + 2 x - 3 ) - cos ( x ) ( 8 x + 2 )}{( 4 x ^{2} + 2 x - 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{- cos ( x )}{4 x ^{2} + 2 x - 3})

Simplify

\frac{- cos ( x )}{4 x ^{2} + 2 x - 3} : - \frac{cos ( x )}{4 x ^{2} + 2 x - 3}

= \frac{d}{d x} (- \frac{cos ( x )}{4 x ^{2} + 2 x - 3})

Take the constant out:

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

= - \frac{d}{d x} (\frac{cos ( x )}{4 x ^{2} + 2 x - 3})

Apply the Quotient Rule:

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

= - \frac{\frac{d}{d x} ( cos ( x ) ) ( 4 x ^{2} + 2 x - 3 ) - \frac{d}{d x} ( 4 x ^{2} + 2 x - 3 ) cos ( x )}{( 4 x ^{2} + 2 x - 3 ) ^{2}}

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

\frac{d}{d x} (4 x^{2} + 2 x - 3) = 8 x + 2

= - \frac{( - sin ( x ) ) ( 4 x ^{2} + 2 x - 3 ) - ( 8 x + 2 ) cos ( x )}{( 4 x ^{2} + 2 x - 3 ) ^{2}}

Simplify

= - \frac{- sin ( x ) ( 4 x ^{2} + 2 x - 3 ) - cos ( x ) ( 8 x + 2 )}{( 4 x ^{2} + 2 x - 3 ) ^{2}}

\frac{\partial}{\partial y} (ln (x + y^{3}))

\frac{d y}{d t} = 4 y

n = 1 \sum \infty \frac{n + 1}{2 n + 1}

y = \frac{5}{4 x + 1}

y = x^{3 \cdot 7}

What is the derivative of (-cos(x))/(4x^2+2x-3) ?
The derivative of (-cos(x))/(4x^2+2x-3) is -(-sin(x)(4x^2+2x-3)-cos(x)(8x+2))/((4x^2+2x-3)^2)
What is the first derivative of (-cos(x))/(4x^2+2x-3) ?
The first derivative of (-cos(x))/(4x^2+2x-3) is -(-sin(x)(4x^2+2x-3)-cos(x)(8x+2))/((4x^2+2x-3)^2)