How do you find the derivative of #sinx/x#?

Question

Evelyn Agard · Accepted Answer

You can use the quotient rule or the product rule. Quotient Rule
#h(x) = f(x)/g(x)#
#h'(x) = (g(x)*f'(x) - f(x)*g'(x))/((g(x))^2)# #h(x) = sinx/x#
#h'(x) = (x*(sinx)' - sinx*(x)')/(x^2) = (xcosx - sinx)/x^2#
#= cosx/x - sinx/x^2# Product Rule
#h(x) = f(x)*g(x)#
#h(x) = f(x)*g'(x) + g(x)*f'(x)# #h(x) = 1/x*sinx#
#h'(x) = 1/x*(sinx)' + sinx*(1/x)' = cosx/x - (sinx)/x^2# As you can see, you still get the same result in the end.

Evelyn Agard · Answer

undefined To find the derivative of $ \frac{{\sin(x)}}{x} $, you can use the quotient rule. The quotient rule states that for functions $ u(x) $ and $ v(x) $, the derivative of $ \frac{{u(x)}}{v(x)} $ is given by:

$$ \left( \frac{{u(x)}}{v(x)} \right)' = \frac{{u'(x)v(x) - u(x)v'(x)}}{(v(x))^2} $$

Applying this rule to $ \frac{{\sin(x)}}{x} $, where $ u(x) = \sin(x) $ and $ v(x) = x $, and noting that the derivative of $ \sin(x) $ is $ \cos(x) $, and the derivative of $ x $ is $ 1 $, the derivative of $ \frac{{\sin(x)}}{x} $ is:

$$ \frac{{d}}{{dx}}\left(\frac{{\sin(x)}}{x}\right) = \frac{{(x \cdot \cos(x) - \sin(x) \cdot 1)}}{x^2} $$

Simplifying this expression gives:

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