What is the derivative of cos(3x^2+4)*6x ?

The derivative of cos(3x^2+4)*6x is 6(-6x^2sin(3x^2+4)+cos(3x^2+4))

What is the first derivative of cos(3x^2+4)*6x ?

The first derivative of cos(3x^2+4)*6x is 6(-6x^2sin(3x^2+4)+cos(3x^2+4))

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derivative of

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

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

Solution steps

\frac{d}{d x} (cos (3 x^{2} + 4) \cdot 6 x)

Take the constant out:

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

= 6 \frac{d}{d x} (cos (3 x^{2} + 4) x)

Apply the Product Rule:

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

f = cos (3 x^{2} + 4), g = x

= 6 (\frac{d}{d x} (cos (3 x^{2} + 4)) x + \frac{d x}{d x} cos (3 x^{2} + 4))

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

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

= 6 ((- sin (3 x^{2} + 4) \cdot 6 x) x + 1 \cdot cos (3 x^{2} + 4))

Simplify

6 ((- sin (3 x^{2} + 4) \cdot 6 x) x + 1 \cdot cos (3 x^{2} + 4)) : 6 (- 6 x^{2} sin (3 x^{2} + 4) + cos (3 x^{2} + 4))

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

\int \frac{1}{3 x ^{2} + 6 x + 6} d x

sin (x), x = 0, y = 0, x = \frac{π}{6}

y^{^{''}} + 4 y^{^{'}} + 4 y = x \cdot cos (2 x) + 3 e^{- 2 x}

\frac{d x}{d t} + 3 x = e^{- 2 t}

x \to 0 - lim ((1 + x)^{\frac{1}{x}})

What is the derivative of cos(3x^2+4)*6x ?
The derivative of cos(3x^2+4)*6x is 6(-6x^2sin(3x^2+4)+cos(3x^2+4))
What is the first derivative of cos(3x^2+4)*6x ?
The first derivative of cos(3x^2+4)*6x is 6(-6x^2sin(3x^2+4)+cos(3x^2+4))