What is the derivative of y=(4x-1)^2(x^4+1)^2 ?

The derivative of y=(4x-1)^2(x^4+1)^2 is 160x^9-72x^8+8x^7+192x^5-80x^4+8x^3+32x-8

What is the first derivative of y=(4x-1)^2(x^4+1)^2 ?

The first derivative of y=(4x-1)^2(x^4+1)^2 is 160x^9-72x^8+8x^7+192x^5-80x^4+8x^3+32x-8

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derivative of

y = (4 x - 1)^{2} (x^{4} + 1)^{2}

160 x^{9} - 72 x^{8} + 8 x^{7} + 192 x^{5} - 80 x^{4} + 8 x^{3} + 32 x - 8

Solution steps

\frac{d}{d x} ((4 x - 1)^{2} (x^{4} + 1)^{2})

Apply the Product Rule:

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

f = (4 x - 1)^{2}, g = (x^{4} + 1)^{2}

= \frac{d}{d x} ((4 x - 1)^{2}) (x^{4} + 1)^{2} + \frac{d}{d x} ((x^{4} + 1)^{2}) (4 x - 1)^{2}

\frac{d}{d x} ((4 x - 1)^{2}) = 8 (4 x - 1)

\frac{d}{d x} ((x^{4} + 1)^{2}) = 8 x^{3} (x^{4} + 1)

= 8 (4 x - 1) (x^{4} + 1)^{2} + 8 x^{3} (x^{4} + 1) (4 x - 1)^{2}

Simplify

8 (4 x - 1) (x^{4} + 1)^{2} + 8 x^{3} (x^{4} + 1) (4 x - 1)^{2} : 160 x^{9} - 72 x^{8} + 8 x^{7} + 192 x^{5} - 80 x^{4} + 8 x^{3} + 32 x - 8

= 160 x^{9} - 72 x^{8} + 8 x^{7} + 192 x^{5} - 80 x^{4} + 8 x^{3} + 32 x - 8

\frac{\partial}{\partial x} (\frac{x z + yz}{x y})

f (x) = \frac{x}{2 + x}

\frac{d}{d x} (\frac{A}{x})

x \to 1 lim (x^{2} + 2 x - \frac{5}{2})

(x^{2} + 1) \frac{d y}{d x} + 4 x (y - 1) = 0, y (0) = 8

What is the derivative of y=(4x-1)^2(x^4+1)^2 ?
The derivative of y=(4x-1)^2(x^4+1)^2 is 160x^9-72x^8+8x^7+192x^5-80x^4+8x^3+32x-8
What is the first derivative of y=(4x-1)^2(x^4+1)^2 ?
The first derivative of y=(4x-1)^2(x^4+1)^2 is 160x^9-72x^8+8x^7+192x^5-80x^4+8x^3+32x-8