How do you find the derivative for #f(x)=cotx/sinx#?

Question

Emilia Aguilar · Accepted Answer

#y^' = -(1 + cos^2x)/sin^3x#

You can differentiate this function by using the quotient rule, which tells you that the derivative of afunction expressed as the quotient of two other functions

#f(x) = g(x)/h(x)#

can be found by using

#color(blue)(d/dx(f(x)) = (g^'(x) * h(x) - g(x) * h^'(x))/[g(x)]^2)#, with #h(x)!=0#.

You can simplify this expression by using the trigonometric identity

#cot(x) = 1/tan(x) = cos(x)/sin(x)#

This means that you can write

#f(x) = cosx/sinx * 1/sinx = cosx/sin^2x#

This function's derivative will thus be

#d/dx(f(x)) = ([d/dx(cosx)] * sin^2x - cosx * d/dx(sin^2x))/(sin^2x)^2#

You can use the power and chain rules to find #d/dx(sin^2x)#.

More specifically, you can write #sin^2x = u^2#, with #u = sinx#. This will get you

#d/dx(u^2) = d/du(u^2) * d/dx(sinx)#

#d/dx(u^2) = 2u * cosx#

#d/dx(sin^2x) = 2sinx * cosx#

Plug this back into your target derivative to get

#f^' = (-sinx * sin^2x - cosx * 2 * sinx * cosx)/sin^4x#

#f^' = (-sin^3x - 2sinx cos^2x)/sin^4x#

You can simplify this further by

#f^' = - (color(red)(cancel(color(black)(sinx)))(sin^2x + 2cos^2x))/sin^(color(red)(cancel(color(black)(4))) color(blue)(3))x#

#f^' = -(overbrace(sin^2x + cos^2x)^(color(red)("=1")) + cos^2x)/sin^3x#

#f^' = color(green)(-(1 + cos^2x)/sin^3x)#

Charles Arocha · Answer

undefined To find the derivative of $ f(x) = \frac{\cot x}{\sin x} $, you can use the quotient rule, which states that for functions $ u(x) $ and $ v(x) $:

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

Using this rule, the derivative of $ f(x) $ is:

$$ f'(x) = \frac{(\cot(x))' \cdot (\sin(x)) - (\cot(x)) \cdot (\sin(x))'}{(\sin(x))^2} $$

Differentiating $ \cot(x) $ and $ \sin(x) $:

$$ (\cot(x))' = -\csc^2(x) $$
$$ (\sin(x))' = \cos(x) $$

Substituting these into the derivative formula:

$$ f'(x) = \frac{(-\csc^2(x)) \cdot (\sin(x)) - (\cot(x)) \cdot (\cos(x))}{(\sin(x))^2} $$

$$ f'(x) = \frac{-\csc^2(x)\sin(x) - \cot(x)\cos(x)}{\sin^2(x)} $$