What is the derivative of y=\sqrt[4]{8x^2-9} ?

The derivative of y=\sqrt[4]{8x^2-9} is (4x)/((8x^2-9)^{3/4)}

What is the first derivative of y=\sqrt[4]{8x^2-9} ?

The first derivative of y=\sqrt[4]{8x^2-9} is (4x)/((8x^2-9)^{3/4)}

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derivative of

y = 4 8 x^{2} - 9

\frac{4 x}{( 8 x ^{2} - 9 ) ^{\frac{3}{4}}}

Solution steps

\frac{d}{d x} (4 8 x^{2} - 9)

Apply the chain rule:

\frac{1}{4 ( 8 x ^{2} - 9 ) ^{\frac{3}{4}}} \frac{d}{d x} (8 x^{2} - 9)

= \frac{1}{4 ( 8 x ^{2} - 9 ) ^{\frac{3}{4}}} \frac{d}{d x} (8 x^{2} - 9)

\frac{d}{d x} (8 x^{2} - 9) = 16 x

= \frac{1}{4 ( 8 x ^{2} - 9 ) ^{\frac{3}{4}}} \cdot 16 x

Simplify

\frac{1}{4 ( 8 x ^{2} - 9 ) ^{\frac{3}{4}}} \cdot 16 x : \frac{4 x}{( 8 x ^{2} - 9 ) ^{\frac{3}{4}}}

= \frac{4 x}{( 8 x ^{2} - 9 ) ^{\frac{3}{4}}}

\frac{\partial}{\partial x} (z^{x y} + (x z)^{y})

\frac{d}{d x} ((x^{3} - 2 x^{2} + 3 x + 1)^{11})

\frac{\partial}{\partial y} (\frac{x}{( x ^{2} + y ^{2} ) ^{2}})

2 \frac{d y}{d t} + y = 0, y (0) = - 3

ln (2) \cdot 2^{x^{2} + 2} x

What is the derivative of y=\sqrt[4]{8x^2-9} ?
The derivative of y=\sqrt[4]{8x^2-9} is (4x)/((8x^2-9)^{3/4)}
What is the first derivative of y=\sqrt[4]{8x^2-9} ?
The first derivative of y=\sqrt[4]{8x^2-9} is (4x)/((8x^2-9)^{3/4)}