What is the derivative of sqrt(x^4-5x^2+9) ?

The derivative of sqrt(x^4-5x^2+9) is (x(2x^2-5))/(sqrt(x^4-5x^2+9))

What is the first derivative of sqrt(x^4-5x^2+9) ?

The first derivative of sqrt(x^4-5x^2+9) is (x(2x^2-5))/(sqrt(x^4-5x^2+9))

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\frac{d}{d x} (x^{4} - 5 x^{2} + 9)

\frac{x ( 2 x ^{2} - 5 )}{x ^{4} - 5 x ^{2} + 9}

Solution steps

\frac{d}{d x} (x^{4} - 5 x^{2} + 9)

Apply the chain rule:

\frac{1}{2 x ^{4} - 5 x ^{2} + 9} \frac{d}{d x} (x^{4} - 5 x^{2} + 9)

= \frac{1}{2 x ^{4} - 5 x ^{2} + 9} \frac{d}{d x} (x^{4} - 5 x^{2} + 9)

\frac{d}{d x} (x^{4} - 5 x^{2} + 9) = 4 x^{3} - 10 x

= \frac{1}{2 x ^{4} - 5 x ^{2} + 9} (4 x^{3} - 10 x)

Simplify

\frac{1}{2 x ^{4} - 5 x ^{2} + 9} (4 x^{3} - 10 x) : \frac{x ( 2 x ^{2} - 5 )}{x ^{4} - 5 x ^{2} + 9}

= \frac{x ( 2 x ^{2} - 5 )}{x ^{4} - 5 x ^{2} + 9}

\frac{\partial}{\partial x} (48 x^{3} cos (6 x^{2} y))

\frac{d}{d x} (\frac{x - 3}{x})

x y^{^{'}} + 3 y = \frac{2 e ^{- 4 x}}{x ^{2}}

\int csc^{2} (u) d u

What is the derivative of sqrt(x^4-5x^2+9) ?
The derivative of sqrt(x^4-5x^2+9) is (x(2x^2-5))/(sqrt(x^4-5x^2+9))
What is the first derivative of sqrt(x^4-5x^2+9) ?
The first derivative of sqrt(x^4-5x^2+9) is (x(2x^2-5))/(sqrt(x^4-5x^2+9))