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

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

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

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

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\frac{d}{d 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}}}

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{d}{d u} (u^{2} + v^{2})

n = 1 \sum \infty \frac{( - 3 ) ^{n}}{7 ^{n + 1}}

\int 8 (y^{4} + 4 y^{2} + 1)^{2} (y^{3} + 2 y) d y

y^{^{'}} + 3 (1 + x^{2}) y = 0

\frac{d}{d m} (mi)

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