Your first-line derivative is correct.
g'(x) = (x^2+1)^5[6(x^2+2)^5(2x)] + (x^2+2)^6[5(x^2+1)^4(2x)]
= 12x(x^2+1)^5 (x^2+2)^5 + 10x(x^2+1)^4 (x^2+2)^6
= 2x(x^2+1)^4 (x^2+2)^5 [6(x^2+1) + 5(x^2+2)]
= 2x(x^2+1)^4 (x^2+2)^5 (6x^2 + 6 + 5x^2 + 10)
= 2x(x^2+1)^4 (x^2+2)^5 (11x^2 + 16)
Setting this to zero gives us,
2x = 0
x = 0
All the other factors would have complex roots.
Since critical points are real points on real graphs, any solution would have to come from the set of real numbers.
So x = 0 is the only one that will yield real values
#2.
y = tan (3x)
y' =3sec^2 (3x) or 3(sec(3x))^2
y'' = 6 sec (3x) (3) (3sec(3x) tan(3x) )
= 18 sec^2 (3x) tan(3x)
Differentiate to find critical numbers and leave function in fully factored form.
g(x) = (x^2+1)^5(x^2+2)^6
g'(x) = (x^2+1)^5[6(x^2+2)^5(2x)] + (x^2+2)^6[5(x^2+1)^4(2x)]
g'(x) = 2x(x^2+1)^4(x^2+2)^5[6(x^2+2)(2x) + 5(x^2+1)(2x)]
g'(x) = 2x(x^2+1)^4(x^2+2)^5(11x^2+11)
g'(x) = 22x(x^2+1)^5(x^2+2)^5
Would critical numbers include:
22x = 0 is 0
But would both (x^2+1)^5 and (x^2+2)^5 remove the fifth power, resulting in;
x^2 = -1, and x^2 = -2
Would these be classified as DNE or not critical numbers?
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Second question,
Find the second derivative of f(x) = tan(3x)
f'(x) = sec^2(3x)(3)
f"(x) = [sec^2(3x)]'(3)'
f"(x) = [sec^2(3x)tan^2(3x)](3)
I'm assuming after the first derivative we keep the sec^2(3x) and (3) separate. But would my route actually need the product rule? Am I supposed to keep the 3 in [sec^2(3x)]' to clear up the confusion of (3)' in the second line?
1 answer
1 answer