If dy = 0 then you have a horizontal line.
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m=dy/dx at (0,b)
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
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
You take the change in Y or dy and divide it by the change in X or dx. Slope equals dy/dx.
You have : y = e^(5x)^2 and de^u/dx = [ e^u ] [ du/dx ]dy/dx = [ e^(5x)^2 ] [ 10x ]
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
73.35+4.37+yAdd 4.37 to 73.35 to get 77.72.(d)/(dy) 73.35+4.37+y=77.72+ySince 77.72 does not contain y, the derivative of 77.72 is 0.(d)/(dy) 73.35+4.37+y=0+(d)/(dy) yTo find the derivative of y, multiply the base (y) by the exponent (1), then subtract 1 from the exponent (1-1=0). Since the exponent is now 0, y is eliminated from the term.(d)/(dy) 73.35+4.37+y=0+1Combine all similar expressions.(d)/(dy) 73.35+4.37+y=1The derivative of 73.35+4.37+y is 1.1
Shqiponja me dy krenare
m=dy/dx at (0,b)
I assume you are asking about using differentiation. In this case, where dy-by-dx=0, there is a stationary point on the graph i.e. where the gradient is equal to 0.
It is one which has a function that does not contain the dependent variable. For example, (dy/dx) + y = f(x) is inhomogeneous but (dy/dx) + y = 0 is not. ( f(x) is a function of the independent variable)
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
Sixto Dy was born on August 6, 1934, in Tarlac Province, Philippines.
Paolo Dy's birth name is Paolo Crisostomo Arbiol Dy.
he was assinated
with y=mx+b dy/dx=m d^2.y/dx^2=0 The rate of change is 0
Ax2 - By2 + Cx + Dy + Exy + F = 0 with A and B having opposite signs