How do you differentiate #f(x)=cotx(cscx)#?

Question

Elijah Abram · Accepted Answer

#- csc^(3)(x) - csc(x) cot^(2)(x)#

We have: #f(x) = cot(x) csc(x)#

This function can be differentiated using the "product rule":

#=> f'(x) = (d) / (dx) (cot(x)) cdot csc(x) + (d) / (dx) (csc(x)) cdot cot(x)#

#=> f'(x) = - csc^(2)(x) cdot csc(x) + (- csc(x) cot(x)) cdot cot(x)#

#=> f'(x) = - csc^(3)(x) - csc(x) cot^(2)(x)#

Paisley Johnson · Answer

undefined To differentiate $ f(x) = \cot(x) \cdot \csc(x) $, you can use the product rule of differentiation. The product rule states that if you have two functions $ u(x) $ and $ v(x) $, then the derivative of their product $ u(x) \cdot v(x) $ with respect to $ x $ is given by:

$$ \frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) $$

Applying the product rule to $ f(x) = \cot(x) \cdot \csc(x) $, we have:

$$ u(x) = \cot(x) $$
$$ v(x) = \csc(x) $$

$$ u'(x) = -\csc^2(x) $$
$$ v'(x) = -\cot(x) \csc(x) $$

Substitute these into the product rule formula:

$$ f'(x) = (-\csc^2(x)) \cdot \csc(x) + \cot(x) \cdot (-\cot(x) \csc(x)) $$

Simplify:

$$ f'(x) = -\csc^3(x) - \cot^2(x) \csc(x) $$

This is the derivative of $ f(x) $ with respect to $ x $.