What is the derivative of y=(cos(4x))/(x^2) ?

The derivative of y=(cos(4x))/(x^2) is -(2(2xsin(4x)+cos(4x)))/(x^3)

What is the first derivative of y=(cos(4x))/(x^2) ?

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

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derivative of

y = \frac{cos ( 4 x )}{x ^{2}}

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (cos (4 x)) = - sin (4 x) \cdot 4

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

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

Simplify

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

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

\int (y - 1)^{- 2} d y

\int x (x^{2} + 3)^{3} d x

\frac{d}{d x} (\frac{2 x + 1}{3 - 5 x})

\int (1 - x)^{9} d x

d er i v a t i v e f (x) = \frac{sin ( x )}{cos ( x )}

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