What is the derivative of sqrt(9x-(sin^2(2x))) ?

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

What is the first derivative of sqrt(9x-(sin^2(2x))) ?

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

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derivative of

9 x - (sin^{2} (2 x))

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = arccsc (x)

t an g e n t f (x) = ln (tan (2 x)), at x = \frac{π}{8}

n = 1 \sum \infty \frac{( - 3 ) ^{n}}{n}

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

d er i v a t i v e \frac{x ^{2} + 1}{x}

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