What is the derivative of y=((t^2)/(t^3-4t))^3 ?

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

What is the first derivative of y=((t^2)/(t^3-4t))^3 ?

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

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derivative of

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

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

Solution steps

\frac{d}{d t} ((\frac{t ^{2}}{t ^{3} - 4 t})^{3})

Apply the chain rule:

3 (\frac{t ^{2}}{t ^{3} - 4 t})^{2} \frac{d}{d t} (\frac{t ^{2}}{t ^{3} - 4 t})

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

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

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

Simplify

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

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

\frac{d y}{d x} = y e^{x}

\int - 2 x^{2} (x^{3} + 6)^{4} d x

\int 2 x^{3} + \frac{1}{2} + \frac{1}{4 x ^{3}} d x

\int_{1}^{6} \frac{ln ( x )}{x ^{2}} d x

\int (\frac{1}{x ^{2} + x + 1}) d x

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