How do you find the derivative of $\frac{cos (2 x^{4}) - 1}{x^{7}}$ $[cos(2x^4) - 1]/x^7$ ?

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How do you find the derivative of $\frac{cos (2 x^{4}) - 1}{x^{7}}$ $[cos(2x^4) - 1]/x^7$ ?

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Answer 1 · 2018-03-21T13:45:41

$\frac{d y}{d x} = - \frac{8}{x^{4}} {cos x}^{4} + \frac{14}{x^{8}} {sin}^{2} x^{4}$ $dy/dx=-8/x^4cosx^4+14/x^8sin^2x^4$

Explanation:

Let
$y = \frac{cos (2 x^{4}) - 1}{x^{7}}$ $y=(cos(2x^4)-1)/(x^7)$

Let
$u = cos (2 x^{4}) - 1$ $u=cos(2x^4)-1$
$v = x^{7}$ $v=x^7$

$y = \frac{u}{v}$ $y=u/v$

By quotient rule,

$\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$ $dy/dx=(v(du)/dx-u(dv)/dx)/v^2$

$u = cos (2 x^{4}) - 1$ $u=cos(2x^4)-1$
Let
$t = x^{4}$ $t=x^4$

$u = cos 2 t - 1$ $u=cos2t-1$
wkt
$1 - cos 2 t = 2 {sin}^{2} t$ $1-cos2t=2sin^2t$
$u = - (1 - cos 2 t)$ $u=-(1-cos2t)$
$u = - 2 {sin}^{2} t$ $u=-2sin^2t$
By chain rule
$\frac{d u}{d x} = \frac{d u}{d t} \frac{d t}{d x}$ $(du)/dx=(du)/dt(dt)/dx$

$u = - 2 {sin}^{2} t$ $u=-2sin^2t$
${sin}^{2} t = {(sin t)}^{2}$ $sin^2t=(sint)^2$
Let
$p = sin t$ $p=sint$

${(sin t)}^{2} = p^{2}$ $(sint)^2=p^2$

$u = - 2 p$ $u=-2p$
By chain rule
$\frac{d u}{d t} = - 2 \frac{d p}{d t}$ $(du)/dt=-2(dp)/dt$

$p = sin t$ $p=sint$

$\frac{d p}{d t} = cos t$ $(dp)/dt=cost$

$\frac{d u}{d t} = - 2 cos t$ $(du)/dt=-2cost$

$- 2 cos t = - 2 {cos x}^{4}$ $-2cost=-2cosx^4$

$\frac{d u}{d t} = - 2 {cos x}^{4}$ $(du)/dt=-2cosx^4$

$\frac{d u}{d x} = \frac{d u}{d t} \frac{d t}{d x}$ $(du)/dx=(du)/dt(dt)/dx$

$t = x^{4}$ $t=x^4$

$\frac{d t}{d x} = 4 x^{3}$ $(dt)/dx=4x^3$

$\frac{d u}{d x} = - 2 {cos x}^{4} (4 x^{3})$ $(du)/dx=-2cosx^4(4x^3)$

$\frac{d u}{d x} = - 8 x^{3} {cos x}^{4}$ $(du)/dx=-8x^3cosx^4$

$v = x^{7}$ $v=x^7$

$\frac{d v}{d x} = 7 x^{6}$ $(dv)/dx=7x^6$

$\frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$ $dy/dx=(v(du)/dx-u(dv)/dx)/v^2$

$u = - 2 {sin}^{2} t$ $u=-2sin^2t$
$u = - 2 {sin}^{2} x^{4}$ $u=-2sin^2x^4$
$v = x^{7}$ $v=x^7$
$\frac{d u}{d x} = - 8 x^{3} {cos x}^{4}$ $(du)/dx=-8x^3cosx^4$
$\frac{d v}{d x} = 7 x^{6}$ $(dv)/dx=7x^6$

$\frac{d y}{d x} = \frac{x^{7} (- 8 x^{3} {cos x}^{4}) - (- 2 {sin}^{2} x^{4}) (7 x^{6})}{{(x^{7})}^{2}}$ $dy/dx=(x^7(-8x^3cosx^4)-(-2sin^2x^4)(7x^6))/(x^7)^2$

$\frac{d y}{d x} = \frac{- 8 x^{10} {cos x}^{4} + 14 x^{6} {sin}^{2} x^{4}}{x^{14}}$ $dy/dx=(-8x^10cosx^4+14x^6sin^2x^4)/x^14$

$\frac{d y}{d x} = - \frac{8}{x^{4}} {cos x}^{4} + \frac{14}{x^{8}} {sin}^{2} x^{4}$ $dy/dx=-8/x^4cosx^4+14/x^8sin^2x^4$

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How do you find the derivative of $\frac{cos (2 x^{4}) - 1}{x^{7}}$ $[cos(2x^4) - 1]/x^7$ ?

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