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

The derivative of ((x^3-1/(2x^3+1))^4) is (36x^2(x^3-1)^3)/((2x^3+1)^5)

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

The first derivative of ((x^3-1/(2x^3+1))^4) is (36x^2(x^3-1)^3)/((2x^3+1)^5)

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

\frac{36 x ^{2} ( x ^{3} - 1 ) ^{3}}{( 2 x ^{3} + 1 ) ^{5}}

Solution steps

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

Apply the chain rule:

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

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

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

= 4 (\frac{x ^{3} - 1}{2 x ^{3} + 1})^{3} \frac{9 x ^{2}}{( 2 x ^{3} + 1 ) ^{2}}

Simplify

4 (\frac{x ^{3} - 1}{2 x ^{3} + 1})^{3} \frac{9 x ^{2}}{( 2 x ^{3} + 1 ) ^{2}} : \frac{36 x ^{2} ( x ^{3} - 1 ) ^{3}}{( 2 x ^{3} + 1 ) ^{5}}

= \frac{36 x ^{2} ( x ^{3} - 1 ) ^{3}}{( 2 x ^{3} + 1 ) ^{5}}

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

f (x) = \frac{2}{( 1 - x ) ^{3}}

\frac{\partial}{\partial x} (e^{\frac{x ^{2}}{2}})

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

y^{^{''}} + 3.26 y^{^{'}} + 2.66 y = 0

What is the derivative of ((x^3-1/(2x^3+1))^4) ?
The derivative of ((x^3-1/(2x^3+1))^4) is (36x^2(x^3-1)^3)/((2x^3+1)^5)
What is the first derivative of ((x^3-1/(2x^3+1))^4) ?
The first derivative of ((x^3-1/(2x^3+1))^4) is (36x^2(x^3-1)^3)/((2x^3+1)^5)