What is the derivative of ((x^3-1/(x^3+1))^8) ?

The derivative of ((x^3-1/(x^3+1))^8) is (48x^2(x^3-1)^7)/((x^3+1)^9)

What is the first derivative of ((x^3-1/(x^3+1))^8) ?

The first derivative of ((x^3-1/(x^3+1))^8) is (48x^2(x^3-1)^7)/((x^3+1)^9)

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\frac{d}{d x} ((\frac{x ^{3} - 1}{x ^{3} + 1})^{8})

\frac{48 x ^{2} ( x ^{3} - 1 ) ^{7}}{( x ^{3} + 1 ) ^{9}}

Solution steps

\frac{d}{d x} ((\frac{x ^{3} - 1}{x ^{3} + 1})^{8})

Apply the chain rule:

8 (\frac{x ^{3} - 1}{x ^{3} + 1})^{7} \frac{d}{d x} (\frac{x ^{3} - 1}{x ^{3} + 1})

= 8 (\frac{x ^{3} - 1}{x ^{3} + 1})^{7} \frac{d}{d x} (\frac{x ^{3} - 1}{x ^{3} + 1})

\frac{d}{d x} (\frac{x ^{3} - 1}{x ^{3} + 1}) = \frac{6 x ^{2}}{( x ^{3} + 1 ) ^{2}}

= 8 (\frac{x ^{3} - 1}{x ^{3} + 1})^{7} \frac{6 x ^{2}}{( x ^{3} + 1 ) ^{2}}

Simplify

8 (\frac{x ^{3} - 1}{x ^{3} + 1})^{7} \frac{6 x ^{2}}{( x ^{3} + 1 ) ^{2}} : \frac{48 x ^{2} ( x ^{3} - 1 ) ^{7}}{( x ^{3} + 1 ) ^{9}}

= \frac{48 x ^{2} ( x ^{3} - 1 ) ^{7}}{( x ^{3} + 1 ) ^{9}}

f (x) = 1 + sin (x), at x = π

f (x) = x^{3} - 74 x^{2} + 40 x + 27490

f (x) = sin^{3} (3 x)

f (t) = cos (2 t)

\frac{d}{d x} (2 \cdot cos (x))

What is the derivative of ((x^3-1/(x^3+1))^8) ?
The derivative of ((x^3-1/(x^3+1))^8) is (48x^2(x^3-1)^7)/((x^3+1)^9)
What is the first derivative of ((x^3-1/(x^3+1))^8) ?
The first derivative of ((x^3-1/(x^3+1))^8) is (48x^2(x^3-1)^7)/((x^3+1)^9)