Well, isn't that a happy little question! To find the derivative of x divided by y, you can use the quotient rule. The derivative will be (y * d/dx(x) - x * d/dx(y)) / (y^2). Just remember, there are no mistakes, just happy little accidents in math!
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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
If y = 3x +- 1, the derivative with respect to x is y' = 3.
y = ln (x) dy/dx = 1/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
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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).
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
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
If y = 3x +- 1, the derivative with respect to x is y' = 3.
The derivative with respect to 'x' is 4y3 . The derivative with respect to 'y' is 12xy2 .
There should not be any y in the derivative itself since y or y(x) is the function whose derivative you are finding.
if y=7/(x^3)-4/xthenby the quotient ruley'=(((0)(x^3)-(7(3x^2))/((x^3)^2))-(((0x)-(4))/(x^2))
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).
Y = 36cot(x)Y' = dy/dx36cot(x)= - 36csc2(x)==========
- 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
Given y=ln(1/x) y'=(1/(1/x))(-x-2)=(1/(1/x))(1/x2)=x/x2=1/x Use the chain rule. The derivative of ln(x) is 1/x. Instead of just "x" inside the natural log function, it's "1/x". Since the inside of the function is not x, the derivative must be multiplied by the derivative of the inside of the function. So it's 1/(1/x) [the derivative of the outside function, natural log] times -x-2=1/x2 [the derivative of the inside of the function, 1/x] This all simplifies to 1/x So the derivative of ln(1/x) is 1/x