What is the derivative of arcsec(1/(4t^5)) ?

The derivative of arcsec(1/(4t^5)) is -(20t^4)/(sqrt(1-16t^{10))}

What is the first derivative of arcsec(1/(4t^5)) ?

The first derivative of arcsec(1/(4t^5)) is -(20t^4)/(sqrt(1-16t^{10))}

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derivative of

arcsec (\frac{1}{4 t ^{5}})

- \frac{20 t ^{4}}{1 - 16 t ^{10}}

Solution steps

\frac{d}{d t} (arcsec (\frac{1}{4 t ^{5}}))

Apply the chain rule:

\frac{1}{( \frac{1}{4 t ^{5}} ) ^{2} ( \frac{1}{4 t ^{5}} ) ^{2} - 1} \frac{d}{d t} (\frac{1}{4 t ^{5}})

= \frac{1}{( \frac{1}{4 t ^{5}} ) ^{2} ( \frac{1}{4 t ^{5}} ) ^{2} - 1} \frac{d}{d t} (\frac{1}{4 t ^{5}})

\frac{d}{d t} (\frac{1}{4 t ^{5}}) = - \frac{5}{4 t ^{6}}

= \frac{1}{( \frac{1}{4 t ^{5}} ) ^{2} ( \frac{1}{4 t ^{5}} ) ^{2} - 1} (- \frac{5}{4 t ^{6}})

Simplify

\frac{1}{( \frac{1}{4 t ^{5}} ) ^{2} ( \frac{1}{4 t ^{5}} ) ^{2} - 1} (- \frac{5}{4 t ^{6}}) : - \frac{20 t ^{4}}{1 - 16 t ^{10}}

= - \frac{20 t ^{4}}{1 - 16 t ^{10}}

n = 0 \sum \infty \frac{z ^{n}}{n ^{2}}

\int 2 x^{13} e^{- x^{7}} d x

\int x^{3} 64 + x^{2} d x

\frac{d ^{6}}{d x ^{6}} (sin (x^{2}))

x \to 0 lim (\frac{x}{sin ( 7 x )})

What is the derivative of arcsec(1/(4t^5)) ?
The derivative of arcsec(1/(4t^5)) is -(20t^4)/(sqrt(1-16t^{10))}
What is the first derivative of arcsec(1/(4t^5)) ?
The first derivative of arcsec(1/(4t^5)) is -(20t^4)/(sqrt(1-16t^{10))}