Let z = √(x^2 +2x +4).
Then your expression becomes
y = (z^2)^z
y = z^(2z)
And the derivative with respect to x is
y' = z^(2z)*(2z' +2ln(z)z')
= (2z')(1 +ln(z))(z^(2z))
Now
z' = (2x +2)/(2√(x^2 +2x +4)
= (x +1)/√(x^2 +2x +4)
So, your derivative is
y' = 2(x +1)/√(x^2 +2x +4) * (1 + (1/2)ln(x^2 +2x +4)) * (x^2 +2x +4)^√(x^2 +2x +4)
For x = 0, this becomes
2(1)/√4 * (1 +(1/2)ln(4)) * 4^√4
= 16*(1 + ln(2))
Answer:
46, 55
Step-by-step explanation:
Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just don’t want to waste my time in case you don’t want me to :)
21/2 is 10.5 so that would be the length of the diagonal near the 4 and the side would be 4 which 10.5 is the lengths of the other 2 diagonals so you take LxW
Answer:1.08 words per second
Step-by-step explanation:
see picture :)
Answer:
Step-by-step explanation:
A anything below 5.7