What is the derivative of sqrt(x^4-9x^2+16) ?

The derivative of sqrt(x^4-9x^2+16) is (x(2x^2-9))/(sqrt(x^4-9x^2+16))

What is the first derivative of sqrt(x^4-9x^2+16) ?

The first derivative of sqrt(x^4-9x^2+16) is (x(2x^2-9))/(sqrt(x^4-9x^2+16))

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\frac{d}{d x} (x^{4} - 9 x^{2} + 16)

\frac{x ( 2 x ^{2} - 9 )}{x ^{4} - 9 x ^{2} + 16}

Solution steps

\frac{d}{d x} (x^{4} - 9 x^{2} + 16)

Apply the chain rule:

\frac{1}{2 x ^{4} - 9 x ^{2} + 16} \frac{d}{d x} (x^{4} - 9 x^{2} + 16)

= \frac{1}{2 x ^{4} - 9 x ^{2} + 16} \frac{d}{d x} (x^{4} - 9 x^{2} + 16)

\frac{d}{d x} (x^{4} - 9 x^{2} + 16) = 4 x^{3} - 18 x

= \frac{1}{2 x ^{4} - 9 x ^{2} + 16} (4 x^{3} - 18 x)

Simplify

\frac{1}{2 x ^{4} - 9 x ^{2} + 16} (4 x^{3} - 18 x) : \frac{x ( 2 x ^{2} - 9 )}{x ^{4} - 9 x ^{2} + 16}

= \frac{x ( 2 x ^{2} - 9 )}{x ^{4} - 9 x ^{2} + 16}

\frac{2 s - 1}{s ^{2} ( s + 1 ) ^{3}}

\int \frac{ln ( 4 x )}{x} d x

\frac{d}{d x} (x^{4} e^{5 x})

\int 2 sin (3 x) + 3 sin (2 x) d x

\frac{d}{d x} (ln (cosh (4 x)))

What is the derivative of sqrt(x^4-9x^2+16) ?
The derivative of sqrt(x^4-9x^2+16) is (x(2x^2-9))/(sqrt(x^4-9x^2+16))
What is the first derivative of sqrt(x^4-9x^2+16) ?
The first derivative of sqrt(x^4-9x^2+16) is (x(2x^2-9))/(sqrt(x^4-9x^2+16))