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

The derivative of (x+1^{2/3}(2x^2-1)^3) is (2(2x^2-1)^3+36x(2x^2-1)^2(x+1))/(3(x+1)^{1/3)}

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

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

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

\frac{2}{3 ( x + 1 ) ^{\frac{1}{3}}} (2 x^{2} - 1)^{3} + 12 x (2 x^{2} - 1)^{2} (x + 1)^{\frac{2}{3}} : \frac{2 ( 2 x ^{2} - 1 ) ^{3} + 36 x ( 2 x ^{2} - 1 ) ^{2} ( x + 1 )}{3 ( x + 1 ) ^{\frac{1}{3}}}

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

\frac{d}{d x} (sin (k x))

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

\int 2 u^{2} d u

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

\int sin^{5} (\frac{x}{2}) d x

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