dy/dx=cos x, if y(0)=0?
832532
dy/dx=cos x\n\nIntegrating both sides, we will get:\ny = sinx + c\n\ny(0) = sin0 + c = 0 => c=0\n\nSo y=sinx.
785428
y=e^cos(x) + cos(e^x)\ny'=-sin(x)e^cos(x) + -sin(e^x)e^x\n=-(sin(x)e^cos(x)+sin(e^x)e^x)
589798
y' = 2x - 2\n2x - 2 = 0\n2x = 2\nx = 1
1034892
What is the function? I assume it is \n\n(dy/dx is equal to 1 only if y=x+C whre C is a constant)\n\nTo prove use the definition of the derivative which is dy/dx as a limit of delta_x appraching of (delta_y/delta_x)\nwe have \ny2-y1=x2+C-x1-C then\ndelta_y=x2-X1\n\ndelta_y/delta_x=(x2-x1)/(x2-x1)=1 QED\n(the is no need to take a limit, however in must be used. I hav eommited it for lack of better representation)\n\nBest
732220
dx/dt=1/2\ndy/dt=2t\ndy/dx=dy/dt/dx/dt\n=2t/1/2=4t\nat t=-1 slope=-4\nat t=1 slope=4
754976
It's true.\nThe answer is indeed y = 0.\nYou can find this by dividing each side by -4.\n-4y = 0\n-4y/-4 = 0/-4\ny = 0\nRemember that zero divided by any nonzero number is always zero.
70526
=((-4y)^2)/(10y-5)= (16y^2)/(10y-5)\nand if the equation is equal to zero then y=4i
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
364111
take the first derivative and set it equal to zero and solve.  2x +1 = 0 x= -1/2  solve for y.  y=-13/4\nvertex is at x= -13/4 y=-2
274505
dy/du=6u^2\ndy/du=(2x^2)/3
274298
Use implicit differentiation to find dy/dx. Standard differentiation rules apply, except when you differentiate a term with a "y" in it you must tack on a dy/dx, so:\n\n2x+8y(dy/dx)=0\ndy/dx = -x/4y\n\nThe tangent line intersects the ellipse at (x1,y1), so the slope of the tangent line will be -x1/(4*y1).\n\nWe can also compute the slope of the tangent line by plugging the two points (-5,0) and (x1,y1) into the standard slope formula m = delta(y)/delta(x) and get that the slope of the tangent line is y1/(x1+5). Since these expressions both describe the slope of the tangent line, they must be equal, so\n-x1/(4*y1)=y1/(x1+5)\nx1^2+5*x1=-4*y1^2\nx1^2+4*y1^2+5*x1=0\n\nSince (x1,y1) is a point on the ellipse, it must satisfy the equation of the ellipse, so x1^2+4*y1^2=5. Substituting this in to x1^2+4*y1^2+5*x1=0 gives us 5+5*x1=0, and x1=-1. At this point, there are actually two y1 values that will satisfy the equation of the ellipse, y1 could be either plus or minus 1 (get these by letting x=-1 in the original equation of the ellipse), and so there are actually two points on the ellipse where the tangent line passes through (-5,0), and those points are (-1,1) and (-1,-1).
399118
y-int= (0,-5)\nx-int= (-1.2,0)
1134427
12x + 4y = 8\n(d/dx) * (12x + 4y = 8)\n12 + 4dy/dx = 0\ndy/dx = -12/4\ndy/dx = - 3
880796
dy/dx = √(y/x) = √y/√x\nFirst of all, group √x,dx and √y,dy to get:\n(1/√y)dy = (1/√x)dx \ny^(-1/2)dy = x^(-1/2)dx\n\n∫ y^(-1/2)dy = ∫ x^(-1/2)dx\n2y^(1/2) = 2x^(1/2) + C1\ny^(1/2) = x^(1/2) + C\ny = (√x + C)²
945033
Derivative is y=cos x -5. Substituting in 0 gives you the slope -4. Now plug in the coordinates (0,3) to get the line, which is y=-4x+3
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
195576
Y = U + VX^2\nVX^2 = Y - U\nX^2 = (Y-U)/V\nX = sqrt ( (Y-U)/V )