What is the derivative of y= 1/((t^4+1)^3) ?

The derivative of y= 1/((t^4+1)^3) is -(12t^3)/((t^4+1)^4)

What is the first derivative of y= 1/((t^4+1)^3) ?

The first derivative of y= 1/((t^4+1)^3) is -(12t^3)/((t^4+1)^4)

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derivative of

y = \frac{1}{( t ^{4} + 1 ) ^{3}}

- \frac{12 t ^{3}}{( t ^{4} + 1 ) ^{4}}

Solution steps

\frac{d}{d t} (\frac{1}{( t ^{4} + 1 ) ^{3}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

= - \frac{3}{( t ^{4} + 1 ) ^{4}} \cdot 4 t^{3}

Simplify

- \frac{3}{( t ^{4} + 1 ) ^{4}} \cdot 4 t^{3} : - \frac{12 t ^{3}}{( t ^{4} + 1 ) ^{4}}

= - \frac{12 t ^{3}}{( t ^{4} + 1 ) ^{4}}

\frac{\partial}{\partial y} (\frac{2 x ^{2} y}{2 + x ^{2} y ^{2}})

n = 0 \sum \infty \frac{1}{ln ( n + 2 )}

\int \frac{x}{3 x ^{2} + 4} d x

x \to + 3 lim (2 x^{2} - 3 x + 4)

\int \frac{1}{x ^{\frac{5}{3}}} d x

What is the derivative of y= 1/((t^4+1)^3) ?
The derivative of y= 1/((t^4+1)^3) is -(12t^3)/((t^4+1)^4)
What is the first derivative of y= 1/((t^4+1)^3) ?
The first derivative of y= 1/((t^4+1)^3) is -(12t^3)/((t^4+1)^4)