What is the derivative of y=(2tan^2(2x)-1)^{1/2} ?

The derivative of y=(2tan^2(2x)-1)^{1/2} is (4sec^2(2x)tan(2x))/((2tan^2(2x)-1)^{1/2)}

What is the first derivative of y=(2tan^2(2x)-1)^{1/2} ?

The first derivative of y=(2tan^2(2x)-1)^{1/2} is (4sec^2(2x)tan(2x))/((2tan^2(2x)-1)^{1/2)}

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derivative of

y = (2 tan^{2} (2 x) - 1)^{\frac{1}{2}}

\frac{4 sec ^{2} ( 2 x ) tan ( 2 x )}{( 2 tan ^{2} ( 2 x ) - 1 ) ^{\frac{1}{2}}}

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{2 ( 2 tan ^{2} ( 2 x ) - 1 ) ^{\frac{1}{2}}} \cdot 8 sec^{2} (2 x) tan (2 x)

Simplify

\frac{1}{2 ( 2 tan ^{2} ( 2 x ) - 1 ) ^{\frac{1}{2}}} \cdot 8 sec^{2} (2 x) tan (2 x) : \frac{4 sec ^{2} ( 2 x ) tan ( 2 x )}{( 2 tan ^{2} ( 2 x ) - 1 ) ^{\frac{1}{2}}}

= \frac{4 sec ^{2} ( 2 x ) tan ( 2 x )}{( 2 tan ^{2} ( 2 x ) - 1 ) ^{\frac{1}{2}}}

\frac{\partial}{\partial x} (x (1 - x - 2 y))

y^{'} = 0.002 y (2500 - y)

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

π^{2}

\int \frac{2}{3 x - 2} d x

What is the derivative of y=(2tan^2(2x)-1)^{1/2} ?
The derivative of y=(2tan^2(2x)-1)^{1/2} is (4sec^2(2x)tan(2x))/((2tan^2(2x)-1)^{1/2)}
What is the first derivative of y=(2tan^2(2x)-1)^{1/2} ?
The first derivative of y=(2tan^2(2x)-1)^{1/2} is (4sec^2(2x)tan(2x))/((2tan^2(2x)-1)^{1/2)}