What is the derivative of y=sqrt(4+csc(t^4)) ?

The derivative of y=sqrt(4+csc(t^4)) is -(2t^3cot(t^4)csc(t^4))/(sqrt(4+csc(t^4)))

What is the first derivative of y=sqrt(4+csc(t^4)) ?

The first derivative of y=sqrt(4+csc(t^4)) is -(2t^3cot(t^4)csc(t^4))/(sqrt(4+csc(t^4)))

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derivative of

y = 4 + csc (t^{4})

- \frac{2 t ^{3} cot ( t ^{4} ) csc ( t ^{4} )}{4 + csc ( t ^{4} )}

Solution steps

\frac{d}{d t} (4 + csc (t^{4}))

Apply the chain rule:

\frac{1}{2 4 + csc ( t ^{4} )} \frac{d}{d t} (4 + csc (t^{4}))

= \frac{1}{2 4 + csc ( t ^{4} )} \frac{d}{d t} (4 + csc (t^{4}))

\frac{d}{d t} (4 + csc (t^{4})) = - 4 t^{3} cot (t^{4}) csc (t^{4})

= \frac{1}{2 4 + csc ( t ^{4} )} (- 4 t^{3} cot (t^{4}) csc (t^{4}))

Simplify

\frac{1}{2 4 + csc ( t ^{4} )} (- 4 t^{3} cot (t^{4}) csc (t^{4})) : - \frac{2 t ^{3} cot ( t ^{4} ) csc ( t ^{4} )}{4 + csc ( t ^{4} )}

= - \frac{2 t ^{3} cot ( t ^{4} ) csc ( t ^{4} )}{4 + csc ( t ^{4} )}

y = (\frac{x - 1}{x ^{2} + x + 1})^{4}

y^{^{'}} + 2 x y = 2 x

\int 5 e^{x} 1 + e^{x} d x

x \to - 9 lim (\frac{x ^{2} - 3}{9 - x})

\int_{0}^{\frac{π}{2}} 12 cos^{5} (x) d x

What is the derivative of y=sqrt(4+csc(t^4)) ?
The derivative of y=sqrt(4+csc(t^4)) is -(2t^3cot(t^4)csc(t^4))/(sqrt(4+csc(t^4)))
What is the first derivative of y=sqrt(4+csc(t^4)) ?
The first derivative of y=sqrt(4+csc(t^4)) is -(2t^3cot(t^4)csc(t^4))/(sqrt(4+csc(t^4)))