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

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

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

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

Popular >

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}}}

\int \frac{cos ( θ )}{2 + sin ^{2} ( θ )} d θ

f (- 7) = \frac{1}{x + 3}

y = \frac{2}{3 x ^{2}}

x \to 0 lim (\frac{4 x ^{2}}{x})

y^{^{'}} = 9 y^{2} + x y^{2}, y (0) = 1

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