Question

Find the first and second derivative of this function using the quotient rule.
y= (x³ + 7) / x

Answer 1

Let\[u=x^3+7\]and\[v=x\]

Answer 2

\[dv/dx=1\]\[du/dx=3x^2+7\]

Answer 3

sorry. du/dx=3x^2

Answer 4

so, you'll get \[(2x^3-7)\over(x^2)\]

Answer 5

for second derivative, let u =2x^3 - 7 and v = x^2

Answer 6

so after doing the same steps, you'll get \[(2x^4+14x)\over(x^4)\]

Answer 7

show your work on that last step.....everything up til that point makes sense

Answer 8

okay. so\[u=2x^3-7\]then\[du/dx=6x^2\]for\[v=x^2\]you'll get\[dv/dx=2x\]

Answer 9

yeah whoops....ok 6x² i was using 6x this entire time....no wonder i spent an hour on this problem

Answer 10

\[x^2(6x^2)-2x(2x^3-7)\over{x^4}\]

Answer 11

oh. haha. yeahh. i always had mistakes like that last time too (: