What is the derivative of y=x^2(4x^4+5x^3-16) ?

The derivative of y=x^2(4x^4+5x^3-16) is 24x^5+25x^4-32x

What is the first derivative of y=x^2(4x^4+5x^3-16) ?

The first derivative of y=x^2(4x^4+5x^3-16) is 24x^5+25x^4-32x

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derivative of

y = x^{2} (4 x^{4} + 5 x^{3} - 16)

24 x^{5} + 25 x^{4} - 32 x

Solution steps

\frac{d}{d x} (x^{2} (4 x^{4} + 5 x^{3} - 16))

Apply the Product Rule:

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

f = x^{2}, g = 4 x^{4} + 5 x^{3} - 16

= \frac{d}{d x} (x^{2}) (4 x^{4} + 5 x^{3} - 16) + \frac{d}{d x} (4 x^{4} + 5 x^{3} - 16) x^{2}

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

\frac{d}{d x} (4 x^{4} + 5 x^{3} - 16) = 16 x^{3} + 15 x^{2}

= 2 x (4 x^{4} + 5 x^{3} - 16) + (16 x^{3} + 15 x^{2}) x^{2}

Simplify

2 x (4 x^{4} + 5 x^{3} - 16) + (16 x^{3} + 15 x^{2}) x^{2} : 24 x^{5} + 25 x^{4} - 32 x

= 24 x^{5} + 25 x^{4} - 32 x

\frac{\partial}{\partial y} (1 + x^{3} + y^{3})

\int_{1}^{4} \frac{2}{x ^{3}} d x

f (x) = 5 cot (\frac{x}{8})

x \to 0 lim (x^{5} ln (x))

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

What is the derivative of y=x^2(4x^4+5x^3-16) ?
The derivative of y=x^2(4x^4+5x^3-16) is 24x^5+25x^4-32x
What is the first derivative of y=x^2(4x^4+5x^3-16) ?
The first derivative of y=x^2(4x^4+5x^3-16) is 24x^5+25x^4-32x