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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))
Y = 36cot(x)Y' = dy/dx36cot(x)= - 36csc2(x)==========
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).
- 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
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.