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Q: The derivative of y x2 - 3x at the point?

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The derivative of y = sin(3x + 5) is 3cos(3x + 5) but only if x is measured in radians.

In this case, you'll need to apply the chain rule, first taking the derivative of the tan function, and multiplying by the derivative of 3x: y = tan(3x) ∴ dy/dx = 3sec2(3x)

Gradient to the curve at any point is the derivative of y = x2 So the gradient is d/dx of x2 = 2x. When x = 2, 2x = 4 so the gradient of the tangent at x = 2 is 4.

What point is a solution to y>3x-2

1 and 2

Related questions

If y = 3x +- 1, the derivative with respect to x is y' = 3.

The derivative of y = sin(3x + 5) is 3cos(3x + 5) but only if x is measured in radians.

3x3 - x2y + 12x - 4y = x2*(3x - y) + 4*(3x - y) = (x2 + 4)*(3x - y)

The derivative of a curve is basically the slope of the curve. If we say, for example, that if y = 2x, the derivative is 2, that means that at any point the line has this slope. If we say that for the function y = x2, the derivative is 2x, that means that at any point "x", the slope is twice the value of "x".

If you mean: y =(lnx)3 then: dy/dx = [3(lnx)2]/x ddy/dx = [(6lnx / x) - 3(lnx)2] / x2 If you mean: y = ln(x3) Then: dy/dx = 3x2/x3 = 3/x = 3x-1 ddy/dx = -3x-2 = -3/x2

y = x2 - 3x - 20 is a formula linking two variables, x and y. It does not - indeed, it cannot - have a solution. The equation x2 - 3x - 20 = 0 has solutions at x = -3.217 and x = 6.217 approx.

x2 + 3y = 7 3x + y2 = 3 3y = x2 + 7 y2 = -3x + 3 y = x2/3 + 7/3 y = ± √(-3x + 3) If you draw the graphs of y = x2/3 + 7/3 and y = ± √(-3x + 3) in a graphing calculator, you will see that they don't intersect, so that the system of the given equations has not a solution.

It is the parabola such that the coordinates of each point on it satisfies the given equation.

In this case, you'll need to apply the chain rule, first taking the derivative of the tan function, and multiplying by the derivative of 3x: y = tan(3x) ∴ dy/dx = 3sec2(3x)

the answers is d

is a quadratic equation for y, in terms of x.

Gradient to the curve at any point is the derivative of y = x2 So the gradient is d/dx of x2 = 2x. When x = 2, 2x = 4 so the gradient of the tangent at x = 2 is 4.

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