What is the derivative of y=(ln(4)x^2)(-x^3-4) ?

The derivative of y=(ln(4)x^2)(-x^3-4) is 2ln(2)(-5x^4-8x)

What is the first derivative of y=(ln(4)x^2)(-x^3-4) ?

The first derivative of y=(ln(4)x^2)(-x^3-4) is 2ln(2)(-5x^4-8x)

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derivative of

y = (ln (4) x^{2}) (- x^{3} - 4)

2 ln (2) (- 5 x^{4} - 8 x)

Solution steps

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

Simplify

ln (4) x^{2} (- x^{3} - 4) : 2 ln (2) x^{2} (- x^{3} - 4)

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

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{2}, g = - x^{3} - 4

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

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

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

= 2 ln (2) (2 x (- x^{3} - 4) + (- 3 x^{2}) x^{2})

Simplify

2 ln (2) (2 x (- x^{3} - 4) + (- 3 x^{2}) x^{2}) : 2 ln (2) (- 5 x^{4} - 8 x)

= 2 ln (2) (- 5 x^{4} - 8 x)

y = x + 12, y = x^{2}

\int \frac{1}{x ^{2} 121 - x ^{2}} d x

x^{2} - 1, - x^{2}

t \to 0 lim (cos (2 t))

44 x + \frac{1584}{x ^{2}}

What is the derivative of y=(ln(4)x^2)(-x^3-4) ?
The derivative of y=(ln(4)x^2)(-x^3-4) is 2ln(2)(-5x^4-8x)
What is the first derivative of y=(ln(4)x^2)(-x^3-4) ?
The first derivative of y=(ln(4)x^2)(-x^3-4) is 2ln(2)(-5x^4-8x)