How do you find the first and second derivatives of #f(x)=cos(x)/x^8 # using the quotient rule?

Question

Paisley Johnson · Accepted Answer

First derivative: #-\frac{x\sin(x)+8\cos(x)}{x^9}#
Second derivative: #\frac{7\sin(x)-x\cos(x)}{x^9}-\frac{9(\sin(x)-8\cos(x))}{x^10}# The quotient #\frac[f(x)}{g(x)}# of two derivable functions #f(x)# and #g(x)# is derivable for all #x# in which #g(x) e 0#. And its derivative is equal to: #frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{(g(x))^2}# So, going back to the exercise we have: #f(x)=\sin(x)# #g(x)=x^8# Thus: #frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{x^8\cdot (-\sin(x)) - \cos(x)\cdot 8x^7}{(x^8)^2}# #frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{-x^8\sin(x)-8x^7\cos(x)}{x^16}# Factoring: #frac{d}{dx}(\frac[f(x)}{g(x)})=\frac{-x^7(x\sin(x)+8\cos(x))}{x^16}# Reducing #frac{x^7}{x^16}# we finally get: #frac{d}{dx}(\frac[f(x)}{g(x)})=-\frac{x\sin(x)+8\cos(x)}{x^9}# For the second derivative your functions #f(x)# and #g(x)# will be: #f(x)=-(x\sin(x)+8\cos(x))# #g(x)=x^9# Try doing it!

Ezekiel Alexander · Answer

undefined To find the first derivative of $ f(x) = \frac{\cos(x)}{x^8} $ using the quotient rule:

1. Apply the quotient rule: $ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2} $
2. Let $ u = \cos(x) $ and $ v = x^8 $.
3. Find the derivatives $ u' $ and $ v' $: 
   - $ u' = -\sin(x) $
   - $ v' = 8x^7 $
4. Substitute into the quotient rule formula:
   $ f'(x) = \frac{x^8(-\sin(x)) - \cos(x)(8x^7)}{(x^8)^2} $
   $ f'(x) = \frac{-x^8\sin(x) - 8x^7\cos(x)}{x^{16}} $
   $ f'(x) = \frac{-\sin(x)x^8 - 8x^7\cos(x)}{x^{16}} $

To find the second derivative, differentiate $ f'(x) $ using the quotient rule again:

1. Apply the quotient rule:
   $ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{vu' - uv'}{v^2} $
2. Let $ u = -\sin(x)x^8 - 8x^7\cos(x) $ and $ v = x^{16} $.
3. Find the derivatives $ u' $ and $ v' $: 
   - $ u' = -\cos(x)x^8 - 8x^7\sin(x) - 56x^6\cos(x) $
   - $ v' = 16x^{15} $
4. Substitute into the quotient rule formula:
   $ f''(x) = \frac{x^{16}(-\cos(x)x^8 - 8x^7\sin(x) - 56x^6\cos(x)) - (-\sin(x)x^8 - 8x^7\cos(x))(16x^{15})}{(x^{16})^2} $
   $ f''(x) = \frac{-x^{24}\cos(x) - 8x^{23}\sin(x) - 56x^{22}\cos(x) + 128x^{22}\cos(x) + 128x^{22}\sin(x)}{x^{32}} $
   $ f''(x) = \frac{-x^{24}\cos(x) - 8x^{23}\sin(x) - 56x^{22}\cos(x) + 128x^{22}\cos(x) + 128x^{22}\sin(x)}{x^{32}} $