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

The derivative of (tan(x)^{1/x}) is tan^{1/x}(x)(-(ln(tan(x))}{x^2}+\frac{2csc(2x))/x)

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

The first derivative of (tan(x)^{1/x}) is tan^{1/x}(x)(-(ln(tan(x))}{x^2}+\frac{2csc(2x))/x)

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

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

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

(tan (x))^{\frac{1}{x}} = e^{\frac{1}{x} l n (t a n (x))}

= \frac{d}{d x} (e^{\frac{1}{x} l n (t a n (x))})

Apply the chain rule:

e^{\frac{1}{x} l n (t a n (x))} \frac{d}{d x} (\frac{1}{x} ln (tan (x)))

= e^{\frac{1}{x} l n (t a n (x))} \frac{d}{d x} (\frac{1}{x} ln (tan (x)))

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

= e^{\frac{1}{x} l n (t a n (x))} (- \frac{ln ( tan ( x ) )}{x ^{2}} + \frac{2 csc ( 2 x )}{x})

e^{\frac{1}{x} l n (t a n (x))} = tan^{\frac{1}{x}} (x)

= tan^{\frac{1}{x}} (x) (- \frac{ln ( tan ( x ) )}{x ^{2}} + \frac{2 csc ( 2 x )}{x})

\int \frac{1}{x ^{2} + 5 x - 6} d x

\frac{d}{d x} (2 c x)

\int \frac{1}{x ( ln ( x ^{4} ) ) ^{9}} d x

e^{x} y \frac{d y}{d x} = e^{- y} + e^{- 2 x - y}

3 sin (2 t)

What is the derivative of (tan(x)^{1/x}) ?
The derivative of (tan(x)^{1/x}) is tan^{1/x}(x)(-(ln(tan(x))}{x^2}+\frac{2csc(2x))/x)
What is the first derivative of (tan(x)^{1/x}) ?
The first derivative of (tan(x)^{1/x}) is tan^{1/x}(x)(-(ln(tan(x))}{x^2}+\frac{2csc(2x))/x)