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.
y = x^5 + x^3 + x^2 + x y' = 5x^(5 - 1) + 3x^(3 - 1) + 2x^(2 -1) + 1x^(1 - 1) (note x^1 = x and x^0 = 1) y' = 5x^4 + 3x^2 + 2x + 1
If you mean x2-3x+5 then the answer is none because its discriminant is less than zero
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)
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
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".
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.
f(x) = 1/3x3 f'(x) = ∫ 1/3x3 dx f'(x) = 1/3∫ x3 dx f'(x) = 1/3 * 3 * x2 f'(x) = x2
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.
the answers is d
is a quadratic equation for y, in terms of x.