What is (\partial)/(\partial x)((x^4+3x^2)(1-5x)) ?

The answer to (\partial)/(\partial x)((x^4+3x^2)(1-5x)) is -25x^4+4x^3-45x^2+6x

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(\partial)/(\partial x)((x^4+3x^2)(1-5x))

\frac{\partial}{\partial x} ((x^{4} + 3 x^{2}) (1 - 5 x))

- 25 x^{4} + 4 x^{3} - 45 x^{2} + 6 x

Solution steps

\frac{\partial}{\partial x} ((x^{4} + 3 x^{2}) (1 - 5 x))

Apply the Product Rule:

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

f = x^{4} + 3 x^{2}, g = 1 - 5 x

= \frac{\partial}{\partial x} (x^{4} + 3 x^{2}) (1 - 5 x) + \frac{\partial}{\partial x} (1 - 5 x) (x^{4} + 3 x^{2})

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

\frac{\partial}{\partial x} (1 - 5 x) = - 5

= (4 x^{3} + 6 x) (1 - 5 x) + (- 5) (x^{4} + 3 x^{2})

Simplify

(4 x^{3} + 6 x) (1 - 5 x) + (- 5) (x^{4} + 3 x^{2}) : - 25 x^{4} + 4 x^{3} - 45 x^{2} + 6 x

= - 25 x^{4} + 4 x^{3} - 45 x^{2} + 6 x

y^{^{''}} - 6 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 1

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

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

\int e^{x^{3} - 3 x^{2}} d y

x \to 5 lim (\frac{6 x}{x - 5})

What is (\partial)/(\partial x)((x^4+3x^2)(1-5x)) ?
The answer to (\partial)/(\partial x)((x^4+3x^2)(1-5x)) is -25x^4+4x^3-45x^2+6x