What is derivative of 1-x^3=t^2 ?

The answer to derivative of 1-x^3=t^2 is (dx)/(dt)=-(2t)/(3x^2)

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derivative of 1-x^3=t^2

derivative of

1 - x^{3} = t^{2}

\frac{d x}{d t} = - \frac{2 t}{3 x ^{2}}

Solution steps

1 - x^{3} = t^{2}

Treat

x

x (t)

Differentiate both sides:

- 3 x^{2} \frac{d x}{d t} = 2 t

Isolate

\frac{d x}{d t} : \frac{d x}{d t} = - \frac{2 t}{3 x ^{2}}

\frac{d x}{d t} = - \frac{2 t}{3 x ^{2}}

(x + y)^{2} = 4

x^{2} - y^{2} = tan (x) y

x = \frac{y + 1}{y}

2 x^{3} + 4 x^{2} - y = x^{6}

y + 5 i = (2 x + i) (x + 2 i)

What is derivative of 1-x^3=t^2 ?
The answer to derivative of 1-x^3=t^2 is (dx)/(dt)=-(2t)/(3x^2)