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?
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)
To find the critical numbers of a function, you need to set the derivative of the function equal to zero and solve for x.
For the function g(x) = (x^2+1)^5(x^2+2)^6, we already have the derivative g'(x) = 22x(x^2+1)^5(x^2+2)^5.
To find the critical numbers, we need to set g'(x) equal to zero and solve for x:
22x(x^2+1)^5(x^2+2)^5 = 0
Here, we have three factors: 22x, (x^2+1)^5, and (x^2+2)^5. To satisfy the equation, either one or more of these factors must be equal to zero.
From 22x = 0, we can solve for x and get x = 0. So, x = 0 is a critical number.
Now, let's look at the remaining factors: (x^2+1)^5 and (x^2+2)^5. To determine if they can be equal to zero, we set each factor equal to zero and solve for x:
(x^2+1)^5 = 0
This equation has no real solutions because a squared term added to 1 will always be positive.
(x^2+2)^5 = 0
This equation also has no real solutions for the same reason.
Therefore, the critical number for this function is x = 0. The factors (x^2+1)^5 and (x^2+2)^5 do not contribute to the critical numbers because they cannot be equal to zero.
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Now, let's address the second question about finding the second derivative of f(x) = tan(3x).
To find the second derivative, we first find the first derivative of f(x), which you correctly calculated as f'(x) = sec^2(3x)(3).
Now, to find the second derivative f''(x), we take the derivative of f'(x) with respect to x. Here, we can apply the chain rule.
f''(x) = [sec^2(3x)]'(3x)' * (3x)'
To clarify the notation, let's denote u = 3x. We have:
f''(x) = [sec^2(u)]'(u)' * (u)'
Now, we can see that we need to use the product rule to differentiate [sec^2(u)]'(u) with respect to u. Applying the product rule, we have:
[sec^2(u)]'(u) = 2sec(u)sec(u)tan(u) = 2sec^2(u)tan(u)
Substituting this back into the equation for f''(x), we get:
f''(x) = [2sec^2(u)tan(u)] * (u)'
Remember that u = 3x. So, we have:
f''(x) = 2sec^2(3x)tan(3x) * (3x)'
Since (u)' is the derivative of u with respect to x, and u = 3x, we know that (u)' equals 3. Therefore, we can simplify the equation further:
f''(x) = 2sec^2(3x)tan(3x) * 3
Finally, we have the second derivative of f(x) as:
f''(x) = 6sec^2(3x)tan(3x)
So, the second derivative is 6 times sec^2(3x)tan(3x).