What is the ((sin(x))/x)^' ?

The ((sin(x))/x)^' is (xcos(x)-sin(x))/(x^2)

What is the first ((sin(x))/x)^' ?

The first ((sin(x))/x)^' is (xcos(x)-sin(x))/(x^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\frac{sin ( x )}{x})^{^{'}}

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

Solution steps

(\frac{sin ( x )}{x})^{'}

Apply the Quotient Rule:

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

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

(sin (x))^{'} = cos (x)

x^{'} = 1

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

Simplify

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

a re a 9 - x^{2}, 0

\int \frac{7 x ^{4} - 6 x ^{2} - x + 5}{x + 2} d x

\frac{d}{d x} (sin^{2} (2 b x))

\frac{d}{d x} (ln ((2 x + 5)^{3}))

\frac{d}{d x} (arcsech (x))

What is the ((sin(x))/x)^' ?
The ((sin(x))/x)^' is (xcos(x)-sin(x))/(x^2)
What is the first ((sin(x))/x)^' ?
The first ((sin(x))/x)^' is (xcos(x)-sin(x))/(x^2)