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

The derivative of f(x)=cos(x^2)-2x^2sin(x^2) is -6xsin(x^2)-4x^3cos(x^2)

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

The first derivative of f(x)=cos(x^2)-2x^2sin(x^2) is -6xsin(x^2)-4x^3cos(x^2)

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derivative of

f (x) = cos (x^{2}) - 2 x^{2} sin (x^{2})

- 6 x sin (x^{2}) - 4 x^{3} cos (x^{2})

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

Simplify

- sin (x^{2}) \cdot 2 x - 2 (2 x sin (x^{2}) + 2 x^{3} cos (x^{2})) : - 6 x sin (x^{2}) - 4 x^{3} cos (x^{2})

= - 6 x sin (x^{2}) - 4 x^{3} cos (x^{2})

\frac{\partial}{\partial y} ((x + y) (x - y))

\frac{d}{d x} (tan^{40} (sec (cos (x))))

\frac{d}{d x} (x^{e} + \frac{1}{x ^{10}})

- \frac{x}{8 y}

\frac{d ^{2}}{d x ^{2}} ((4 x^{2} + 8 x) cos (x))

What is the derivative of f(x)=cos(x^2)-2x^2sin(x^2) ?
The derivative of f(x)=cos(x^2)-2x^2sin(x^2) is -6xsin(x^2)-4x^3cos(x^2)
What is the first derivative of f(x)=cos(x^2)-2x^2sin(x^2) ?
The first derivative of f(x)=cos(x^2)-2x^2sin(x^2) is -6xsin(x^2)-4x^3cos(x^2)