Taking
and differentiating both sides with respect to
yields
![\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B3x%5E2%2By%5E2%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B7%5Cbigg%5D%5Cimplies%206x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D0)
Solving for the first derivative, we have
Differentiating again gives
![\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B6x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B0%5Cbigg%5D%5Cimplies%206%2B2%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%2B2y%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D0)
Solving for the second derivative, we have
Now, when
and
, we have
Answer:
I'm sorry I would help you I don't understand the second sentence.
Step-by-step explanation:
Answer:
2 and 2/3
Step-by-step explanation:
(2/3)x(2/3)x6=
2 and 2/3
Answer:
42
Step-by-step explanation:
39+39 = 78
162 - 78 = 2 lengths =84
84/2 = 42
The choices are supposed to be
f(x) = sin x + 3
f(x) = cos x + 3
f(x) = 3 sin x
f(x) = 3 cos x
The amplitude is the value of the numerical coefficient of sin or cos. The only possible answers are
f(x) = 3 sin x
f(x) = 3 cos x
Next, the function must pass through the point (0,3)
3 sin 0 = 0 and
3 cos = 3
Therefore, the answer is
f(x) = 3 cos x<span />
