What is the derivative of y=arcsec(1/(7t^3)) ?

The derivative of y=arcsec(1/(7t^3)) is -(21t^2)/(sqrt(1-49t^6))

What is the first derivative of y=arcsec(1/(7t^3)) ?

The first derivative of y=arcsec(1/(7t^3)) is -(21t^2)/(sqrt(1-49t^6))

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derivative of

y = arcsec (\frac{1}{7 t ^{3}})

- \frac{21 t ^{2}}{1 - 49 t ^{6}}

Solution steps

\frac{d}{d t} (arcsec (\frac{1}{7 t ^{3}}))

Apply the chain rule:

\frac{1}{( \frac{1}{7 t ^{3}} ) ^{2} ( \frac{1}{7 t ^{3}} ) ^{2} - 1} \frac{d}{d t} (\frac{1}{7 t ^{3}})

= \frac{1}{( \frac{1}{7 t ^{3}} ) ^{2} ( \frac{1}{7 t ^{3}} ) ^{2} - 1} \frac{d}{d t} (\frac{1}{7 t ^{3}})

\frac{d}{d t} (\frac{1}{7 t ^{3}}) = - \frac{3}{7 t ^{4}}

= \frac{1}{( \frac{1}{7 t ^{3}} ) ^{2} ( \frac{1}{7 t ^{3}} ) ^{2} - 1} (- \frac{3}{7 t ^{4}})

Simplify

\frac{1}{( \frac{1}{7 t ^{3}} ) ^{2} ( \frac{1}{7 t ^{3}} ) ^{2} - 1} (- \frac{3}{7 t ^{4}}) : - \frac{21 t ^{2}}{1 - 49 t ^{6}}

= - \frac{21 t ^{2}}{1 - 49 t ^{6}}

y = ∣ x - 3 ∣, y = \frac{x}{2}

f (x) = tan (sec (x))

y^{^{''}} + 14 y^{^{'}} + 49 y = 0

y^{^{'}} = a y

f (x) = 6

What is the derivative of y=arcsec(1/(7t^3)) ?
The derivative of y=arcsec(1/(7t^3)) is -(21t^2)/(sqrt(1-49t^6))
What is the first derivative of y=arcsec(1/(7t^3)) ?
The first derivative of y=arcsec(1/(7t^3)) is -(21t^2)/(sqrt(1-49t^6))