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

The answer to (\partial)/(\partial x)((4x+y)^3(x-y)^2) is 12(4x+y)^2(x-y)^2+2(x-y)(4x+y)^3

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

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

12 (4 x + y)^{2} (x - y)^{2} + 2 (x - y) (4 x + y)^{3}

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

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

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

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

\frac{\partial}{\partial x} ((4 x + y)^{3}) = 12 (4 x + y)^{2}

\frac{\partial}{\partial x} ((x - y)^{2}) = 2 (x - y)

= 12 (4 x + y)^{2} (x - y)^{2} + 2 (x - y) (4 x + y)^{3}

\frac{d y}{d x} = 14 x^{- 8}

\frac{d}{d x} (x^{3} - 3 x^{2} - 9 x + 15)

x^{3} + 3 y - x y^{^{'}} = 0

g (t) = - \frac{3}{2 t ^{\frac{3}{2}}}

x \to \frac{π}{2} lim (tan^{2} (x))

What is (\partial)/(\partial x)((4x+y)^3(x-y)^2) ?
The answer to (\partial)/(\partial x)((4x+y)^3(x-y)^2) is 12(4x+y)^2(x-y)^2+2(x-y)(4x+y)^3