(a) work the problem as written applying the quotient rule to each fraction,
(b) rewrite f(x) using negative exponents and then find the derivative,
(c) rewrite f(x) as a single fraction and then find the derivative, and
(d) verify that the answers for (a) - (c) are equivalent.
f(x)=10/x^4 + 3/x^2
f(x)=10/x^4 + 3/x^2
1. quotient rule ........ y = f(x)/g(x) → y ' = [ f'(x)g(x) - g'(x)f(x)] / [ g(x)]^2
[0*x^4 - 4x^3 *10] / [x^8] + [0*x^2 - 2x *3]/ [ x^4] =
[ -40x^3] / [x^8] + [ -6x]/ [x^4] =
-40 / x^5 - 6 / x^3
2. f(x) = 10x^-4 + 3x^-2 ....use the power rule.....
f ' (x) = -40x^-5 - 6x^-3 = -40/x^5 - 6/x^3
3. as a single fraction = f(x) = [10 + 3x^2] / x^4
f ' (x) = ( 6x * x^4 - 4x^3 [10 + 3x^2] ) / x^8 =
(6x^5 - 40x^3 - 12x^5) /x^8 =
[-6x^5 - 40x^3] / x^8 =
[-6x^5]/ x^8 - [40x^3] / x^8 =
-6/x^3 - 40/x^5 = -40/x^5 - 6/x^3
4. All are the same
Note, Yura_chan......I rarely [if ever] use the quotient rule......much each easier to write the expression with negative exponents and use the power rule.....IMHO......!!!!
f(x)=10/x^4 + 3/x^2
1. quotient rule ........ y = f(x)/g(x) → y ' = [ f'(x)g(x) - g'(x)f(x)] / [ g(x)]^2
[0*x^4 - 4x^3 *10] / [x^8] + [0*x^2 - 2x *3]/ [ x^4] =
[ -40x^3] / [x^8] + [ -6x]/ [x^4] =
-40 / x^5 - 6 / x^3
2. f(x) = 10x^-4 + 3x^-2 ....use the power rule.....
f ' (x) = -40x^-5 - 6x^-3 = -40/x^5 - 6/x^3
3. as a single fraction = f(x) = [10 + 3x^2] / x^4
f ' (x) = ( 6x * x^4 - 4x^3 [10 + 3x^2] ) / x^8 =
(6x^5 - 40x^3 - 12x^5) /x^8 =
[-6x^5 - 40x^3] / x^8 =
[-6x^5]/ x^8 - [40x^3] / x^8 =
-6/x^3 - 40/x^5 = -40/x^5 - 6/x^3
4. All are the same
Note, Yura_chan......I rarely [if ever] use the quotient rule......much each easier to write the expression with negative exponents and use the power rule.....IMHO......!!!!