What is dy/dx, if y = ( 3x -2)/ (5x^2 +4)?
823232
y = (3x - 2)/(5x² + 4)\n\nW.w.t.f dy/dx.  You can use the formula:\nd/dx (u/v) = (v du/dx - u dv/dx)/v²\n\nNow, we have u = 3x - 2, v = 5x² + 4.  du/dx = 3 and dv/dx = 10x. We can now substitute them to the formula:\ndy/dx = [(5x² + 4)(3) - (3x - 2)(10x)]/(5x² + 4)²\n\nNow,\ndy/dx = (15x² + 12 - 30x² + 20x)/(5x² + 4)²\n\nThus,\ndy/dx = (-15x² + 20x + 12)/(5x² + 4)²\n\ntherefore,\ndy/dx = -(15x² - 20x - 12)/(5x² + 4)²\n\n^_^
702852
equation of the tangent of the line is the derivative of the line\n\ndy/dx = 3x^2 + 2
349210
dy/dx = -(15cos(3/x)/2x^2) * (sin(3/x))^3/2
152532
y = 4x - 3 and y = 5x - 1, so\n4x - 3 = 5x - 1 (I substituted the value of y in the first eq'n to the second eq'n)\n-x = 2\nx = -2\ny = 4(-2) - 3\ny = -8 - 3\ny = -11\n;-)
979217
y = sqrt(1+t^2) = (1+t^2)^(1/2)\n\nby the power and chain rule\n\ndy/dt = (1/2)(1+t^2)^(-1/2) x d/dt (1+ t^2)\n\n= 2t(1/2) /sqrt(1+t^2)\n\n= t/sqrt(1+t^2)\n\nTherefore\n\ndy = (t / sqrt(1+t^2)) dt
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
948239
xy = y^½\n y = y^½ /x\n y = y^½*x^-½\n\ndy/dx = -½ y^½ x^ -3/2\ndy/dx = - [½ y^½ ]/x^+3/2\ndy/dx = - (y^½) / (2x^3/2)
1103883
Y=X^5-X^-3+X^2\nY'=5X^4+3X^-4+2X\nY''=20X^3-12X^-5+2
120944
dy/dx = -4x \n\nAt x=2, gradient is -8\n\ny - y(1) = dy/dx(x - x(1)) [x(1), y(1) means x,y coordinates]\n\ny - 1 = -8*(x - 2)\n\ny = -8x + 17
1049778
y=f(x)=((x-5)/(2x+1))^3\nIf y=f(x)^n then    :    y'=n*f(x)^(n-1)*f'(x)\nNow here:\ny' = 3 * ((x-5)/(2x+1))^2 * [(1*(2x+1)-2*(x-5)) / (2x+1)^2]\nSimplifying:\n\ny'=33*(x-5)^2/(2x+1)^4
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
965203
y = (ln x²)^(2x+3) = e^((2x+3) ln ln x²)\nNow just ruthlessly apply the chain rule:\ndy/dx = e^((2x+3) ln ln x²) * (2 ln ln x² + (2x+3) * 1/ln x² * 1/x² * 2x)\nAnd simplify:\n(2 ln x)^(2x+3) * (2 ln (2 ln x) + (2x+3)/(x ln x))
867402
first evaluate dy/dx = 2x +4\nnow put x=1 \ndy/dx = 2*1 + 4= 6
71733
Take y on the other side\n\ny^2 dy/dx + y^3 = e^(x)\nMultiply by 3\n\n3 y^2 dy/dx + 3 y^3 = 3 e^(x)\n\nFirst term is d/dx (y^3)\nHence put z = y^3 and teh equation becomes\ndz/dx + 3z = 3 e^(x)\n\nNow this is in standard form\nMultiply throughout by e^(3x)\n\ne^(3x) dz/dx+ 3 e^(3x) z = 3 e^(4x)\nLHS can be re-written as\n\nd/dx { e^(3x)* z) = 3 e^(4x)\nIntegrate both sides\ne(3x)*z = 3 e^(4x)/4 +C\n\nz = 3/4 e^(4x)*e^(-3x) + C e^(-3x)\n\nSubstitute back for z=y^3
866423
-2y = 2x^2 - 3x - 4\ny = -x^2 + (3/2)x + 2\nUsing Calculus\ny' = -2x + 3/2\ny' = -2(-2) + 3/2\ny' = -5/2 (slope)\ny=mx+b\n-5 = (-5/2)(-2) + b\n-5 = 5 + b\nb = -10\ny = (-5/2)x - 10
714041
10x+4y+4\n\nor can be written as y= -2.5x-1
850195
I think you mean dy/dx, not dy/dt.\nOK, so we have dy/dx = 18yx^5.\nMove all the terms involving y to one side, and the xs to the other side.\nSo (1/y) dy = 18x^5 dx.\nNow we can integrate both sides:\nintegral of 1/y dy = integral of 18x^5 dx\nln|y| = 3x^6 + c\nNow, to get rid of the ln take e of both sides:\ny = e^(3x^6 + c). You can also write that as y = Ae^(3x^6), because e^c is just a constant - its more usual to write it this way.\nNow substitute in the point - x = 0, y = 5 to get 5 = A, so A = 5.\nThus the equation is y = 5e^(3x^6).