Find the d^3z/dydx^2 partial derivative of:
ln[sin(x-y)]
The answer is supposed to be 1/2; however, I cannot figure out how to get this answer. Any help is appreciated.
I got the dz/dy derivative to be:
cos(x-y)/sin(x-y)
I got the d^2z/dydx derivative to be:
-(sin(x-y))^2 - (cos(x-y))^2/ (sin(x-y))^4
I think there's a minus sign missing there.
∂z/∂y = -cos(x-y)/sin(x-y) = -cot(x-y)
∂^2z/∂y∂x = csc^2(x-y)
∂^3z/∂y∂x^2 = 2csc(x-y) (-csc(x-y)cot(x-y))
= -2csc^2(x-y)cot(x-y)
To find the partial derivative d^3z/dxdy^2, we need to take the partial derivatives successively.
The given function is z = ln[sin(x-y)].
Step 1: Find dz/dx
To find dz/dx, we use the chain rule. Let u = sin(x-y).
dz/dx = dz/du * du/dx
To find dz/du, we differentiate ln(u) with respect to u, which gives us 1/u.
dz/du = 1/u
To find du/dx, we differentiate sin(x-y) with respect to its argument, which is (x-y). So, du/dx = cos(x-y).
Applying the chain rule:
dz/dx = dz/du * du/dx = 1/u * cos(x-y)
Step 2: Find dz/dy
To find dz/dy, we use the chain rule. Let v = sin(x-y).
dz/dy = dz/dv * dv/dy
To find dz/dv, we differentiate ln(v) with respect to v, which gives us 1/v.
dz/dv = 1/v
To find dv/dy, we differentiate sin(x-y) with respect to its argument, which is (x-y). So, dv/dy = -cos(x-y) (since the derivative of -cos(x) is sin(x)).
Applying the chain rule:
dz/dy = dz/dv * dv/dy = 1/v * -cos(x-y)
Step 3: Find d^3z/dydx^2
Finally, we find d^3z/dydx^2 by taking the partial derivative of dz/dx with respect to y.
d^3z/dydx^2 = d/dy(dz/dx) = d/dy[1/u * cos(x-y)]
Using the product rule of differentiation, we have:
d^3z/dydx^2 = d(1/u)/dy * cos(x-y) + 1/u * d(cos(x-y))/dy
To simplify, we need to find du/dy and d^2x/dy^2:
du/dy = -(cos(x-y)) (since the derivative of -cos(x) is sin(x))
d^2x/dy^2 = -sin(x-y)
Plugging in these values:
d^3z/dydx^2 = (-1/u * sin(x-y) * cos(x-y)) + (1/u * du/dy * cos(x-y))
Substituting u = sin(x-y):
d^3z/dydx^2 = (-1/sin(x-y) * sin(x-y) * cos(x-y)) + (1/sin(x-y) * -(cos(x-y)) * cos(x-y))
Simplifying the expression:
d^3z/dydx^2 = (-1 * cos(x-y)) + (1 * -(cos(x-y)))
d^3z/dydx^2 = -2 * cos(x-y)
Therefore, the partial derivative d^3z/dydx^2 = -2 * cos(x-y).
The answer you mentioned, 1/2, may not be correct based on the calculation above. Please verify the correctness of the given answer.
To find the third derivative of ln[sin(x-y)] with respect to y and the second derivative with respect to x, we'll need to apply the chain rule and the product rule multiple times. Here's the step-by-step process:
Step 1: Start by finding the first derivative with respect to y.
Since the function is ln[sin(x-y)], its derivative with respect to y can be found using the chain rule.
d/dy [ln[sin(x-y)]] = (1/sin(x-y)) * d/dy[sin(x-y)]
Step 2: Simplify the expression further by applying the chain rule again.
d/dy [ln[sin(x-y)]] = (1/sin(x-y)) * cos(x-y) * d/dy [x-y]
Step 3: Since we're only interested in the third derivative with respect to y, we'll continue differentiating with respect to y.
d^2/dy^2 [ln[sin(x-y)]] = (1/sin(x-y)) * cos(x-y) * d^2/dy^2 [x-y]
Step 4: Differentiate once more with respect to y.
d^3/dy^3 [ln[sin(x-y)]] = (1/sin(x-y)) * cos(x-y) * d^3/dy^3 [x-y]
Step 5: Now, let's find the third derivative of (x-y) with respect to y.
d^3/dy^3 [x-y] = 0
Since (x-y) does not contain y, its derivative with respect to y is zero.
Step 6: Substitute the third derivative of (x-y) into the expression we obtained in Step 4.
d^3/dy^3 [ln[sin(x-y)]] = (1/sin(x-y)) * cos(x-y) * 0
Step 7: Simplify the expression.
d^3/dy^3 [ln[sin(x-y)]] = 0
Step 8: Finally, we can conclude that the third derivative of ln[sin(x-y)] with respect to y and the second derivative with respect to x is zero.
Therefore, the answer is not 1/2, but rather 0.