What is the derivative of (a^2-x^3/(b^2)) ?

The derivative of (a^2-x^3/(b^2)) is -(3x^2)/(b^2)

What is the first derivative of (a^2-x^3/(b^2)) ?

The first derivative of (a^2-x^3/(b^2)) is -(3x^2)/(b^2)

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\frac{d}{d x} (\frac{a ^{2} - x ^{3}}{b ^{2}})

- \frac{3 x ^{2}}{b ^{2}}

Solution steps

\frac{d}{d x} (\frac{a ^{2} - x ^{3}}{b ^{2}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{b ^{2}} \frac{d}{d x} (a^{2} - x^{3})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{b ^{2}} (\frac{d}{d x} (a^{2}) - \frac{d}{d x} (x^{3}))

\frac{d}{d x} (a^{2}) = 0

\frac{d}{d x} (x^{3}) = 3 x^{2}

= \frac{1}{b ^{2}} (0 - 3 x^{2})

Simplify

\frac{1}{b ^{2}} (0 - 3 x^{2}) : - \frac{3 x ^{2}}{b ^{2}}

= - \frac{3 x ^{2}}{b ^{2}}

x = 7 y^{2}, x = 8 + 5 y^{2}

\frac{2 x}{x + 2}

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

x e^{y} d y + \frac{x ^{2} + 1}{y} d x = 0

\frac{d}{d x} (\frac{1 - 2 x}{1 + 2 x})

What is the derivative of (a^2-x^3/(b^2)) ?
The derivative of (a^2-x^3/(b^2)) is -(3x^2)/(b^2)
What is the first derivative of (a^2-x^3/(b^2)) ?
The first derivative of (a^2-x^3/(b^2)) is -(3x^2)/(b^2)