What is the derivative of sqrt(tan^2(3x)+1) ?

The derivative of sqrt(tan^2(3x)+1) is (3sec^2(3x)tan(3x))/(sqrt(sec^2(3x)))

What is the first derivative of sqrt(tan^2(3x)+1) ?

The first derivative of sqrt(tan^2(3x)+1) is (3sec^2(3x)tan(3x))/(sqrt(sec^2(3x)))

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derivative of

tan^{2} (3 x) + 1

\frac{3 sec ^{2} ( 3 x ) tan ( 3 x )}{sec ^{2} ( 3 x )}

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{2 tan ^{2} ( 3 x ) + 1} \cdot 6 sec^{2} (3 x) tan (3 x)

Simplify

\frac{1}{2 tan ^{2} ( 3 x ) + 1} \cdot 6 sec^{2} (3 x) tan (3 x) : \frac{3 sec ^{2} ( 3 x ) tan ( 3 x )}{sec ^{2} ( 3 x )}

= \frac{3 sec ^{2} ( 3 x ) tan ( 3 x )}{sec ^{2} ( 3 x )}

ln (1 + \frac{1}{x ^{2} + 1})

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

\frac{d}{d x} (1.1 x + 6.3 x^{3})

\frac{d y}{d x} + 3 x y^{2} = 0

\int \frac{3 x ^{4} - 2 x ^{3} - 2 x ^{2} + 2 x - 5}{x ^{3} - x ^{2}} d x

What is the derivative of sqrt(tan^2(3x)+1) ?
The derivative of sqrt(tan^2(3x)+1) is (3sec^2(3x)tan(3x))/(sqrt(sec^2(3x)))
What is the first derivative of sqrt(tan^2(3x)+1) ?
The first derivative of sqrt(tan^2(3x)+1) is (3sec^2(3x)tan(3x))/(sqrt(sec^2(3x)))