How do you differentiate #f(x)= cosx/ (sinx)# twice using the quotient rule?

Question

Christopher Allers · Accepted Answer

#f'(x)=-csc^2x,f''(x)=2csc^2xcotx#

By means of the quotient rule:

#f'(x)=(sinxd/dx[cosx]-cosxd/dx[sinx])/sin^2x#

Remember that:

#d/dx[cosx]=-sinx#

#d/dx[sinx]=cosx#

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

#f'(x)=-(sin^2x+cos^2x)/sin^2x#

#f'(x)=-1/sin^2x#

#f'(x)=-csc^2x#

If we wish to apply the quotient rule once more to find the second derivative, use:

#f'(x)=(-1)/sin^2x#

#f''(x)=(sin^2xd/dx[-1]-(-1)d/dx[sin^2x])/sin^4x#

Once more, identify every derivative:

#d/dx[-1]=0#

Next up, the chain rule:

#d/dx[sin^2x]=2sinxd/dx[sinx]=2sinxcosx#

Reconnect the plug:

#f''(x)=(2sinxcosx)/sin^4x#

#f''(x)=(2cosx)/sin^3x#

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

#f''(x)=2csc^2xcotx#

William Abdalla · Answer

undefined To differentiate $ f(x) = \frac{\cos(x)}{\sin(x)} $ twice using the quotient rule:

1. Find the first derivative $ f'(x) $:
$$ f'(x) = \frac{(\sin(x) \cdot (-\sin(x))) - (\cos(x) \cdot \cos(x))}{(\sin(x))^2} $$

2. Simplify the expression:
$$ f'(x) = \frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)} $$

3. Find the second derivative $ f''(x) $ by differentiating $ f'(x) $ using the quotient rule again:
$$ f''(x) = \frac{[2\sin(x) \cdot \cos(x) \cdot \sin(x)] - [(-\sin^2(x) - \cos^2(x)) \cdot \cos(x)]}{(\sin(x))^4} $$

4. Simplify the expression:
$$ f''(x) = \frac{2\sin^2(x) \cdot \cos(x) + \sin^3(x) + \cos^3(x)}{(\sin(x))^3} $$