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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.
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What is the derivative of x minus y divided by x plus y?
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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 1 divided by x?
Find dy/dx of y=1/x. It may be simpler for you to examine the equation y=x^-1. This equation is the exact same as y=1/x. Therefore just multiply -1 by x and subtract 1 from the exponent giving you -x^-2 or y=-(1/x^2). You can also do it through quotient rules. Therefore take the derivative of the top 1 which = 0 and multiply that by the bottom X which will give you 0. Then subtract the derivative of the bottom x this equals 1 and multiply it by the top (1). Put this all over the bottom squared. Which leads to -1/x^2. y=1/x = y=x^-1 = -x^-2 = -(1/x^2)=dy/dx or y=1/x = ((0*x)-(1*1)/x^2 = dy/dx=-1/x^2
How do you find the derivative of y equals sin8x?
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x
What is the derivative of x minus y divided by x plus y?
m
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 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.
How can I find the derivative of y equals 7 divided by x to the 3rd power subtracted by 4 divided by x?
if y=7/(x^3)-4/xthenby the quotient ruley'=(((0)(x^3)-(7(3x^2))/((x^3)^2))-(((0x)-(4))/(x^2))
What is the derivative of y equals 36cotx?
Y = 36cot(x)Y' = dy/dx36cot(x)= - 36csc2(x)==========
How do you estimate the derivative of a function given a numericallyValues of x and g(x) are given in the table. For what value of x is g'(x) closest to 3?
You can estimate the derivative by looking at adjacent rows of the table, and calculate (difference of y-coordinates) divided by (difference of x-coordinates).
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.
