What is the derivative of y=(tan(x))^{9/x} ?

The derivative of y=(tan(x))^{9/x} is 9tan^{9/x}(x)(-(ln(tan(x))}{x^2}+\frac{2csc(2x))/x)

What is the first derivative of y=(tan(x))^{9/x} ?

The first derivative of y=(tan(x))^{9/x} is 9tan^{9/x}(x)(-(ln(tan(x))}{x^2}+\frac{2csc(2x))/x)

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derivative of

y = (tan (x))^{\frac{9}{x}}

9 tan^{\frac{9}{x}} (x) (- \frac{ln ( tan ( x ) )}{x ^{2}} + \frac{2 csc ( 2 x )}{x})

Solution steps

\frac{d}{d x} ((tan (x))^{\frac{9}{x}})

Apply exponent rule:

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

(tan (x))^{\frac{9}{x}} = e^{\frac{9}{x} l n (t a n (x))}

= \frac{d}{d x} (e^{\frac{9}{x} l n (t a n (x))})

Apply the chain rule:

e^{\frac{9}{x} l n (t a n (x))} \frac{d}{d x} (\frac{9}{x} ln (tan (x)))

= e^{\frac{9}{x} l n (t a n (x))} \frac{d}{d x} (\frac{9}{x} ln (tan (x)))

\frac{d}{d x} (\frac{9}{x} ln (tan (x))) = 9 (- \frac{ln ( tan ( x ) )}{x ^{2}} + \frac{2 csc ( 2 x )}{x})

= e^{\frac{9}{x} l n (t a n (x))} \cdot 9 (- \frac{ln ( tan ( x ) )}{x ^{2}} + \frac{2 csc ( 2 x )}{x})

e^{\frac{9}{x} l n (t a n (x))} = tan^{\frac{9}{x}} (x)

= 9 tan^{\frac{9}{x}} (x) (- \frac{ln ( tan ( x ) )}{x ^{2}} + \frac{2 csc ( 2 x )}{x})

What is the derivative of y=(tan(x))^{9/x} ?
The derivative of y=(tan(x))^{9/x} is 9tan^{9/x}(x)(-(ln(tan(x))}{x^2}+\frac{2csc(2x))/x)
What is the first derivative of y=(tan(x))^{9/x} ?
The first derivative of y=(tan(x))^{9/x} is 9tan^{9/x}(x)(-(ln(tan(x))}{x^2}+\frac{2csc(2x))/x)