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:
B. 1/3 multiplied by k=7
Step-by-step explanation:
7 ÷ ⅓ = k
Multiply both sides by ⅓
7 ÷ ⅓ × ⅓ = k × ⅓
7 = k × ⅓
P1 = (-7, -9)
P2 = (-2, 4)
ratio = 1/4
x = (-7 + 0.25(-2))/(1 + 0.25)
= (-7 - 0.5) / 1.25
= -7.5/1.25
= -6
y = (-9 + 0.25(4)) / (1 + 0.25)
y = (-9 + 1) / 1.25
y = -8/1.25
y = 6.4
The coordinate sof the point are (-6, 6.4)