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What is the derivative of 4xy3?
The derivative with respect to 'x' is 4y3 . The derivative with respect to 'y' is 12xy2 .
What is the y in a derivative of the function yx3-13x2 16x 5?
There should not be any y in the derivative itself since y or y(x) is the function whose derivative you are finding.
What is the derivative of 40xy where x equals 2 and y equals 2?
- the derivative with respect to x is 40y - The derivative with respect to Y is 40xSo, since both x and y equal 2, both derivatives yield 40*2 = 80
What is the derivative of sinx pwr cosx?
For the function: y = sin(x)cos(x) To find the derivative y', implicit differentiation must be used. To do this, both sides of the equation must be put into the argument of a natural logarithm: ln(y) = ln(sin(x)cos(x)) by the properties of logarithms, this can also be expressed as: ln(y) = cos(x)ln(sin(x)) deriving both sides of the equation yields: (1/y)(y') = cos(x)(1/sin(x))(cos(x)) + -sin(x)ln(sin(x)) This derivative features two important things. The obvious thing is the product rule use to differentiate the right side of the equation. The left side of the equation brings into play the "implicit" differentiation part of this problem. The derivative of ln(y) is a chain rule. The derivative of just ln(y) is simply 1/y, but you must also multiply by the derivative of y, which is y'. so the total derivative of ln(y) is (1/y)(y'). solving for y' in the above, the following is found: y' = y[(cos2(x)/sin(x)) - sin(x)ln(sin(x))] = y[cot(x)cos(x) - sin(x)ln(sin(x))] y' = y[cot(x)cos(x) - sin(x)ln(sin(x))] = sin(x)cos(x)[cot(x)cos(x) - sin(x)ln(sin(x)) is the most succinct form of this derivative.
What is a partial derivative?
A partial derivative is the derivative of a function of more than one variable with respect to only one variable. When taking a partial derivative, the other variables are treated as constants. For example, the partial derivative of the function f(x,y)=2x2 + 3xy + y2 with respect to x is:?f/?x = 4x + 3yhere we can see that y terms have been treated as constants when differentiating.The partial derivative of f(x,y) with respect to y is:?f/?y = 3x + 2yand here, x terms have been treated as constants.
What is the derivative of -x-y?
If y is a function of x, that is y=f(x), then the derivative of x-y is 1-y' or 1-dy/dx (where y' or dy/dx is the differential coefficient of y with respect to x).
What is the derivative of x-y?
The partial derivative in relation to x: dz/dx=-y The partial derivative in relation to y: dz/dy= x If its a equation where a constant 'c' is set equal to the equation c = x - y, the derivative is 0 = 1 - dy/dx, so dy/dx = 1
What is the derivative of e the the power ln x?
y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1
What is the derivative for 3x 1?
If y = 3x +- 1, the derivative with respect to x is y' = 3.
What is the derivative of 4xy3?
The derivative with respect to 'x' is 4y3 . The derivative with respect to 'y' is 12xy2 .
What is the derivative of x minus y divided by x plus y?
m
What is the y in a derivative of the function yx3-13x2 16x 5?
There should not be any y in the derivative itself since y or y(x) is the function whose derivative you are finding.
What is the derivative of y equals 36cotx?
Y = 36cot(x)Y' = dy/dx36cot(x)= - 36csc2(x)==========
What is the derivative of x divided by y?
You have to specify more information than that. If y is an independent variable and you're talking about the derivative with respect to x, it would be 1/y.
What is the derivative of 40xy where x equals 2 and y equals 2?
- the derivative with respect to x is 40y - The derivative with respect to Y is 40xSo, since both x and y equal 2, both derivatives yield 40*2 = 80
What is the derivative of sinx pwr cosx?
For the function: y = sin(x)cos(x) To find the derivative y', implicit differentiation must be used. To do this, both sides of the equation must be put into the argument of a natural logarithm: ln(y) = ln(sin(x)cos(x)) by the properties of logarithms, this can also be expressed as: ln(y) = cos(x)ln(sin(x)) deriving both sides of the equation yields: (1/y)(y') = cos(x)(1/sin(x))(cos(x)) + -sin(x)ln(sin(x)) This derivative features two important things. The obvious thing is the product rule use to differentiate the right side of the equation. The left side of the equation brings into play the "implicit" differentiation part of this problem. The derivative of ln(y) is a chain rule. The derivative of just ln(y) is simply 1/y, but you must also multiply by the derivative of y, which is y'. so the total derivative of ln(y) is (1/y)(y'). solving for y' in the above, the following is found: y' = y[(cos2(x)/sin(x)) - sin(x)ln(sin(x))] = y[cot(x)cos(x) - sin(x)ln(sin(x))] y' = y[cot(x)cos(x) - sin(x)ln(sin(x))] = sin(x)cos(x)[cot(x)cos(x) - sin(x)ln(sin(x)) is the most succinct form of this derivative.
How do you find the mean of x over y equals a over b?
Find the derivative of Y and then divide that by the derivative of A
