How do you find the derivative #1/(x^7 + 2)#?

3 Answers
May 24, 2015

I would usr the Quotient Rule as:
#f'(x)=(-7x^6)/(x^7+2)^2#

Remember the Quotient Rule to derive a function #f(x)=(g(x))/(h(x))#
you get:
#f'(x)=(g'(x)h(x)-g(x)h'(x))/[h(x)]^2#

May 24, 2015

To find the derivative of #f(x) = 1/(x^7 + 2)# you would use the quotient rule:

#f(x) = g(x)/(h(x))# then #f'(x) = (g'(x)h(x) - g(x)h'(x))/(h(x)^2)#

Thus:

let #g(x) = 1# so #g'(x) = 0#

let #h(x) = (x^7 + 2)# so #h'(x) = 7x^6#

Therefore we get that:

#f'(x) = (-7x^6)/(x^7 + 2)^2#

this shows us that #f(x)# is always decreasing, as its derivative is always negative

May 25, 2015

I would not use the quotient rule. (There's no difference in the answer, but this is how I think about this differentiation.)

I would use the chain rule:

#f(x)=1/(x^7+2) = (x^7+2)^-1#

#f'(x) = -1(x^7+2)^-2 * 7x^6#

#= (-7x^6)/(x^7+2)^2#