What is the derivative of (sin(x^2-5))/(2x) ?

The derivative of (sin(x^2-5))/(2x) is (2x^2cos(x^2-5)-sin(x^2-5))/(2x^2)

What is the first derivative of (sin(x^2-5))/(2x) ?

The first derivative of (sin(x^2-5))/(2x) is (2x^2cos(x^2-5)-sin(x^2-5))/(2x^2)

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derivative of

\frac{sin ( x ^{2} - 5 )}{2 x}

\frac{2 x ^{2} cos ( x ^{2} - 5 ) - sin ( x ^{2} - 5 )}{2 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{sin ( x ^{2} - 5 )}{2 x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{d}{d x} (\frac{sin ( x ^{2} - 5 )}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( sin ( x ^{2} - 5 ) ) x - \frac{d x}{d x} sin ( x ^{2} - 5 )}{x ^{2}}

\frac{d}{d x} (sin (x^{2} - 5)) = cos (x^{2} - 5) \cdot 2 x

\frac{d x}{d x} = 1

= \frac{1}{2} \cdot \frac{cos ( x ^{2} - 5 ) \cdot 2 xx - 1 \cdot sin ( x ^{2} - 5 )}{x ^{2}}

Simplify

\frac{1}{2} \cdot \frac{cos ( x ^{2} - 5 ) \cdot 2 xx - 1 \cdot sin ( x ^{2} - 5 )}{x ^{2}} : \frac{2 x ^{2} cos ( x ^{2} - 5 ) - sin ( x ^{2} - 5 )}{2 x ^{2}}

= \frac{2 x ^{2} cos ( x ^{2} - 5 ) - sin ( x ^{2} - 5 )}{2 x ^{2}}

x \to 1 lim (\frac{x ^{2} - 1}{x + 1})

\int u cos (2 u) d u

\frac{df}{d x} f (x) = cos (2 x)

(cos (7 x)) y^{^{'}} - 7 (tan (7 x)) y = 9 sec (7 x)

1 - x^{2}, 29 - 11 x

What is the derivative of (sin(x^2-5))/(2x) ?
The derivative of (sin(x^2-5))/(2x) is (2x^2cos(x^2-5)-sin(x^2-5))/(2x^2)
What is the first derivative of (sin(x^2-5))/(2x) ?
The first derivative of (sin(x^2-5))/(2x) is (2x^2cos(x^2-5)-sin(x^2-5))/(2x^2)