What is the derivative of y=2\sqrt[4]{9-x^2} ?

The derivative of y=2\sqrt[4]{9-x^2} is -x/((9-x^2)^{3/4)}

What is the first derivative of y=2\sqrt[4]{9-x^2} ?

The first derivative of y=2\sqrt[4]{9-x^2} is -x/((9-x^2)^{3/4)}

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derivative of

- \frac{x}{( 9 - x ^{2} ) ^{\frac{3}{4}}}

Solution steps

Take the constant out:

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

Apply the chain rule:

\frac{1}{4 ( 9 - x ^{2} ) ^{\frac{3}{4}}} \frac{d}{d x} (9 - x^{2})

= \frac{1}{4 ( 9 - x ^{2} ) ^{\frac{3}{4}}} \frac{d}{d x} (9 - x^{2})

\frac{d}{d x} (9 - x^{2}) = - 2 x

= 2 \cdot \frac{1}{4 ( 9 - x ^{2} ) ^{\frac{3}{4}}} (- 2 x)

Simplify

2 \cdot \frac{1}{4 ( 9 - x ^{2} ) ^{\frac{3}{4}}} (- 2 x) : - \frac{x}{( 9 - x ^{2} ) ^{\frac{3}{4}}}

= - \frac{x}{( 9 - x ^{2} ) ^{\frac{3}{4}}}

x \to 0 lim (- 1 + 2)

\int 6 x^{2} 2 x^{3} + 1 d x

\frac{d y}{d t} = 4 y, y (5) = 9

\int \frac{x + 7}{x ^{2} + 16} d x

x \to - 4 lim (\frac{1}{x + 4})

What is the derivative of y=2\sqrt[4]{9-x^2} ?
The derivative of y=2\sqrt[4]{9-x^2} is -x/((9-x^2)^{3/4)}
What is the first derivative of y=2\sqrt[4]{9-x^2} ?
The first derivative of y=2\sqrt[4]{9-x^2} is -x/((9-x^2)^{3/4)}