What is the derivative of (x^4-1^3(x^3+1)^4) ?

The derivative of (x^4-1^3(x^3+1)^4) is 12x^3(x^4-1)^2(x^3+1)^4+12x^2(x^3+1)^3(x^4-1)^3

What is the first derivative of (x^4-1^3(x^3+1)^4) ?

The first derivative of (x^4-1^3(x^3+1)^4) is 12x^3(x^4-1)^2(x^3+1)^4+12x^2(x^3+1)^3(x^4-1)^3

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\frac{d}{d x} ((x^{4} - 1)^{3} (x^{3} + 1)^{4})

12 x^{3} (x^{4} - 1)^{2} (x^{3} + 1)^{4} + 12 x^{2} (x^{3} + 1)^{3} (x^{4} - 1)^{3}

Solution steps

\frac{d}{d x} ((x^{4} - 1)^{3} (x^{3} + 1)^{4})

Apply the Product Rule:

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

f = (x^{4} - 1)^{3}, g = (x^{3} + 1)^{4}

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

\frac{d}{d x} ((x^{4} - 1)^{3}) = 12 x^{3} (x^{4} - 1)^{2}

\frac{d}{d x} ((x^{3} + 1)^{4}) = 12 x^{2} (x^{3} + 1)^{3}

= 12 x^{3} (x^{4} - 1)^{2} (x^{3} + 1)^{4} + 12 x^{2} (x^{3} + 1)^{3} (x^{4} - 1)^{3}

\int_{1}^{2} \frac{1}{8 - 3 t} d t

f (x) = x^{2} - 1, at x = 0

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

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What is the derivative of (x^4-1^3(x^3+1)^4) ?
The derivative of (x^4-1^3(x^3+1)^4) is 12x^3(x^4-1)^2(x^3+1)^4+12x^2(x^3+1)^3(x^4-1)^3
What is the first derivative of (x^4-1^3(x^3+1)^4) ?
The first derivative of (x^4-1^3(x^3+1)^4) is 12x^3(x^4-1)^2(x^3+1)^4+12x^2(x^3+1)^3(x^4-1)^3