How do you find the derivative of #[cos(2x^4) - 1]/x^7#?

Question

Ezekiel Alexander · Accepted Answer

#dy/dx=-8/x^4cosx^4+14/x^8sin^2x^4#

Let #y=(cos(2x^4)-1)/(x^7)#

Let # u=cos(2x^4)-1# #v=x^7#

#y=u/v#

By quotient rule,

#dy/dx=(v(du)/dx-u(dv)/dx)/v^2#

# u=cos(2x^4)-1# Let #t=x^4#

#u=cos2t-1# wkt #1-cos2t=2sin^2t# #u=-(1-cos2t)# #u=-2sin^2t# By chain rule #(du)/dx=(du)/dt(dt)/dx#

#u=-2sin^2t# #sin^2t=(sint)^2# Let #p=sint#

#(sint)^2=p^2#

#u=-2p# By chain rule #(du)/dt=-2(dp)/dt#

#p=sint#

#(dp)/dt=cost#

#(du)/dt=-2cost#

#-2cost=-2cosx^4#

#(du)/dt=-2cosx^4#

#(du)/dx=(du)/dt(dt)/dx#

#t=x^4#

#(dt)/dx=4x^3#

#(du)/dx=-2cosx^4(4x^3)#

#(du)/dx=-8x^3cosx^4#

#v=x^7#

#(dv)/dx=7x^6#

#dy/dx=(v(du)/dx-u(dv)/dx)/v^2#

# u=-2sin^2t# #u=-2sin^2x^4# #v=x^7# #(du)/dx=-8x^3cosx^4# #(dv)/dx=7x^6#

#dy/dx=(x^7(-8x^3cosx^4)-(-2sin^2x^4)(7x^6))/(x^7)^2#

#dy/dx=(-8x^10cosx^4+14x^6sin^2x^4)/x^14#

#dy/dx=-8/x^4cosx^4+14/x^8sin^2x^4#

Ezekiel Alexander · Answer

undefined To find the derivative of $ \frac{\cos(2x^4) - 1}{x^7} $, you can use the quotient rule, which states that the derivative of $ \frac{u}{v} $ is $ \frac{u'v - uv'}{v^2} $, where $ u $ and $ v $ are functions of $ x $, and $ u' $ and $ v' $ are their respective derivatives with respect to $ x $.

Let $ u = \cos(2x^4) - 1 $ and $ v = x^7 $. Then, $ u' = -8x^3\sin(2x^4) $ and $ v' = 7x^6 $.

Applying the quotient rule:

$$ \frac{d}{dx}\left(\frac{\cos(2x^4) - 1}{x^7}\right) = \frac{(-8x^3\sin(2x^4))(x^7) - (\cos(2x^4) - 1)(7x^6)}{(x^7)^2} $$

Simplifying:

$$ = \frac{-8x^{10}\sin(2x^4) - 7x^6\cos(2x^4) + 7x^6}{x^{14}} $$