1.) differentiate: y=sin (3x/2) + cos (3 x/2)
6.) If 3x3 - 3y3 = 48, find y" at point x = 2.
1.
well, to separate it into two problems (which I would not normally do):
let z = 3 x/2 then dz/dx = 3/2 and dx/dz = 2 /3
y = sin z + cos z
dy/dz = cos z - sin z
but
dy/dx = dy/dz dz/dx = 3/2 [ cos z - sin z]
= 3/2 [ cos 3x/2 -sin 3x/2 ]
2. suspect you mean:
If 3x^3 - 3y^3 = 48, find y" at point x = 2.
y^3 = x^3 - 16
y = (x^3 -16)^(1/3)
dy/dx = (1/3) [(x^3-16)^-(2/3)] (3 x^2)
d/dx(dy/dx) = (1/3) { [(x^3-16)^-(2/3)](6x) + (3x^2)(-2/3)(x^3-16)^-(5/3)(3x^2]}
at x = 2
(1/3) { [(8-16)^-(2/3)](12) + (12)(-2/3)(8-16)^-(5/3)(12]} etc
To differentiate the given function y = sin(3x/2) + cos(3x/2), we can use the chain rule. The chain rule states that if we have a function g(f(x)), where g(u) and f(x) are both differentiable functions, then the derivative of g(f(x)) is given by g'(f(x)) * f'(x).
Applying the chain rule, we start by calculating the derivatives of sin(3x/2) and cos(3x/2).
The derivative of sin(3x/2) can be calculated as follows:
- Take the derivative of the outermost function sin(u), which is cos(u).
- Multiply by the derivative of the inner function (3x/2), which is 3/2.
So, d/dx (sin(3x/2)) = (3/2) * cos(3x/2).
Similarly, the derivative of cos(3x/2) can be calculated as follows:
- Take the derivative of the outermost function cos(u), which is -sin(u).
- Multiply by the derivative of the inner function (3x/2), which is 3/2.
So, d/dx (cos(3x/2)) = (-3/2) * sin(3x/2).
Now, we can find the derivative of y = sin(3x/2) + cos(3x/2) by adding the derivatives of the two terms:
dy/dx = (3/2) * cos(3x/2) + (-3/2) * sin(3x/2).
To find y" at point x = 2, we need to evaluate the second derivative of y at x = 2.
Taking the derivative of the previously obtained expression dy/dx, we apply the chain rule again:
- The derivative of cos(3x/2) is (-3/2) * sin(3x/2).
- The derivative of sin(3x/2) is (3/2) * cos(3x/2).
Thus, the second derivative of y with respect to x is given by:
d^2y/dx^2 = (3/2) * (-3/2) * sin(3x/2) + (3/2) * (3/2) * cos(3x/2).
Now, substitute x = 2 into the expression d^2y/dx^2 to find y" at x = 2:
y" = (3/2) * (-3/2) * sin(3(2)/2) + (3/2) * (3/2) * cos(3(2)/2).
Finally, calculate the numerical value of y" at x = 2.