What is the derivative of sqrt(6+tan^2(x)) ?

The derivative of sqrt(6+tan^2(x)) is (sec^2(x)tan(x))/(sqrt(6+tan^2(x)))

What is the first derivative of sqrt(6+tan^2(x)) ?

The first derivative of sqrt(6+tan^2(x)) is (sec^2(x)tan(x))/(sqrt(6+tan^2(x)))

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\frac{d}{d x} (6 + tan^{2} (x))

\frac{sec ^{2} ( x ) tan ( x )}{6 + tan ^{2} ( x )}

Solution steps

\frac{d}{d x} (6 + tan^{2} (x))

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (\frac{18}{x ^{4}})

\int \frac{1}{sec ^{2} ( u )} d u

\int_{0}^{4} ∣ 2 x - 3 ∣ d x

\frac{s}{s + \frac{1}{2}}

e^{- x}, 5

What is the derivative of sqrt(6+tan^2(x)) ?
The derivative of sqrt(6+tan^2(x)) is (sec^2(x)tan(x))/(sqrt(6+tan^2(x)))
What is the first derivative of sqrt(6+tan^2(x)) ?
The first derivative of sqrt(6+tan^2(x)) is (sec^2(x)tan(x))/(sqrt(6+tan^2(x)))