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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (θ) = sec (θ)

- 0.02 x^{3} + 0.37 x^{2} + 241 x + 17905

(\frac{t}{t + 1 ( t )})^{^{'}}

y^{^{'}} + \frac{y}{x} = \frac{1}{x ^{2}}

y = 4 sin (π x + y), (1, 0)

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