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-(x-1)-1 or -1/(x-1)
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What is the derivative of ln 1 divided by 1-x?
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How do you find the rate of change--or slope--in a two variable equation?
For a straight line, if A = (x1 , y1) and B = (x2 , y2) are any two points on the line, then the slope is (y2 - y1)/(x2 - x1) provided x2 is not the same as x1. More generally, if the equation is y = f(x) then the rate of change in y is dy/dx or f'(x), the derivative of the function f(x).
What if raised to the power of one?
Any number raised to the power 1 is that same number, x1 = x. For example, 51 = 5.
What is the derivative of the square root of 1-sinx?
√(1-sinx)=(1-sinx)1/2Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(1-sinx)1/2=(1/2)(1-sinx)1/2-1*d/dx(1-sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*d/dx(1-sinx)-The derivative of 1-sinx is:d/dx(u-v)=du/dx-dv/dxd/dx(1-sinx)=d/dx(1)-d/dx(sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[d/dx(1)-d/dx(sinx)]-The derivative of 1 is 0 because it is a constant.-The derivative of sinx is:d/dx(sinu)=cos(u)*d/dx(u)d/dx(sinx)=cos(x)*d/dx(x)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*d/dx(x))]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*1)]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x))]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[-cos(x)]d/dx(1-sinx)1/2=(-cosx)/[2√(1-sinx)]
Definition for Sum of Absolute Deviations?
The answer depends on absolute deviation from what: the mean, median or some other measure. Suppose you have n observations, x1, x2, ... xn and you wish to calculate the sum of the absolute deviation of these observations from some fixed number c. The deviation of x1 from c is (x1 - c). The absolute deviation of x1 from c is |x1 - c|. This is the non-negative value of (x1 - c). That is, if (x1 - c) ≤ 0 then |x1 - c| = (x1 - c) while if (x1 - c) < 0 then |(x1 - c)| = - (x1 - c). Then the sum of absolute deviations is the above values, summed over x1, x2, ... xn.
What is the integral of -3x?
In order to compute that integral, we need to use the power rule: ∫ xⁿ dx = xn + 1/(n + 1) + c where n is any constant except 0 and -1. Apply that rule to get: ∫ 3x dx = 3 ∫ x dx [Factor out the constant] = 3 ∫ x1 dx [Make note of the exponent] = 3x1 + 1/(1 + 1) + c = 3x2/2 + c So that is the integral of 3x.
What is the derivative of the square root x?
Use: √x = x1/2 By the Power Rule (Decrease the power by 1. Multiply by the original power.): d/dx √x = d/dx x1/2 = 1/2 x-1/2
Integral of sin square root x?
For ∫ sin(√x) dx let y = √x = x1/2 → dy = 1/2 x-1/2 dx → 2x1/2 dy = dx → 2y dy = dx → ∫ sin(x1/2) dx = ∫(sin y) 2y dy Now: ∫ uv dx = u∫v dx - ∫(u'∫v dx) dx → ∫(sin y) 2y dy = ∫2y sin y dy = 2y ∫sin y dy - ∫(2 ∫sin y dy) dy = -2y cos y + 2 sin y + C = 2 sin y - 2y cos y + C → ∫ sin(√x) dx = 2 sin(√x) - 2(√x) cos(√x) + C
What is differential equation of x-y equals xy?
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
What is the code in c plus plus for line drawing algorithm using OpenGL?
Here is the C code for DDA line drawing... #include<stdio.h> #include<conio.h> #include<graphics.h> #include<math.h> void main() { int gd, gm; float x, y, dx, dy, len; int x1, y1, x2, y2, i; detectgraph(&gd, &gm); initgraph(&gd, &gm,""); printf("\n Enter the coordinates of line : "); scanf("%d %d %d %d", &x1, &y1, &x2, &y2); dx = abs(x2-x1); dy = abs(y2-y1); if (dx >= dy) len = dx; else len = dy; dx = (x2-x1)/len; dy = (y2-y1)/len; x = x1 + 0.5; y = y1 + 0.5; i = 1; while (i <= len) { putpixel(x, y, 6); x = x+dx; y = y+dy; i++; } getch(); } -Suraj.
How can you find the length of arc?
If you want to find the lenght of a curve y = f(x) between two values of x, lets say x1 and x2, you must compute this integral : Intx1 to x2[sqrt(dx2 + dy2)] You can either express the original function in terms of y or in terms of x, but it is much simpler to express it in a way such that the integral will not be improper. For example, lets say we want to find the lenght of arc of the curve y = x2 between x = 0 and x = 1. We could express this function in terms of y but we will keep it this way because if we change it, we will have to compute an improper integral, which can sometimes be very tedious. The differential of y = x2 is dy = 2x dx. We now need to square the differential : (dy)2 = (2x dx)2 = 4x2 (dx)2 We now have to compute this integral: Int0 to 1[sqrt(dx2 + dy2)] = Int0 to 1[sqrt(dx2 + 4x2 dx2)] = Int0 to 1[sqrt(1 + 4x2) dx] This last integral is easy to compute using a trigonometric substitution.
What is the formula for midpoint?
( x1 + x2) divided by 2 then (y1 +y 2) divided by 2
Simplify The Square root of x divided by x?
sqr.rtx/x= sqrt.x*sqr.rtx/sqr.rtx=x/x*sqrt.x=1/sqrt.x. x1/2 = x1/2 * x1/2 = x = 1 (x1/2) /x= 1/x1/2
What is x to the power of one?
x1 = x
X1 plus x2 divided by 2 y1 plus y2 divided by 2?
formula for the midpoint of a line
What is the equation to find slope?
m = y2 - y1 divided by x2 - x1
How do you determine the slope of a line on an graph?
Rise divided by run. (Y2 - Y1) / (X2 - X1) - with (X1, Y1) and (X2, Y2) being two points on the graph.
