What is the derivative of f(x)=-8xsin(4x^2+1) ?

The derivative of f(x)=-8xsin(4x^2+1) is -8(sin(4x^2+1)+8x^2cos(4x^2+1))

What is the first derivative of f(x)=-8xsin(4x^2+1) ?

The first derivative of f(x)=-8xsin(4x^2+1) is -8(sin(4x^2+1)+8x^2cos(4x^2+1))

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derivative of

f (x) = - 8 x sin (4 x^{2} + 1)

- 8 (sin (4 x^{2} + 1) + 8 x^{2} cos (4 x^{2} + 1))

Solution steps

\frac{d}{d x} (- 8 x sin (4 x^{2} + 1))

Take the constant out:

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

= - 8 \frac{d}{d x} (x sin (4 x^{2} + 1))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = sin (4 x^{2} + 1)

= - 8 (\frac{d x}{d x} sin (4 x^{2} + 1) + \frac{d}{d x} (sin (4 x^{2} + 1)) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (sin (4 x^{2} + 1)) = cos (4 x^{2} + 1) \cdot 8 x

= - 8 (1 \cdot sin (4 x^{2} + 1) + cos (4 x^{2} + 1) \cdot 8 xx)

Simplify

- 8 (1 \cdot sin (4 x^{2} + 1) + cos (4 x^{2} + 1) \cdot 8 xx) : - 8 (sin (4 x^{2} + 1) + 8 x^{2} cos (4 x^{2} + 1))

= - 8 (sin (4 x^{2} + 1) + 8 x^{2} cos (4 x^{2} + 1))

\frac{d}{d x} (- f (x) sin (x) + g (x) cos (x))

\int \frac{x + 3}{5 - 4 x - x ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (\frac{( x + b ) ( x - a )}{( x - b )})

(cos (x) sin (x) - x y^{2} 2) d x + y (1 - x^{2}) d y = 0

What is the derivative of f(x)=-8xsin(4x^2+1) ?
The derivative of f(x)=-8xsin(4x^2+1) is -8(sin(4x^2+1)+8x^2cos(4x^2+1))
What is the first derivative of f(x)=-8xsin(4x^2+1) ?
The first derivative of f(x)=-8xsin(4x^2+1) is -8(sin(4x^2+1)+8x^2cos(4x^2+1))