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A particle, P, moves along the x-axis. The displacement, x metres, of P is given by 0.5t^2(t^2 - 2t + 1), when is P instantaneously at rest

differenciate and then eqate with v = 0.

Answered by Harry S. • Maths tutor

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f(x) = sinx. Using differentiation from first principles find the exact value of f' (π/6).

The derivative of the function where x=π/6 is defined asThe limit as h->0 of [sin(h+π/6)-sin(π/6)]/hUsing the double angle formula, sin(h+π/6) = sin(h)cos(π/6) + cos(h)sin(π/6) = √3sin(h)/2 + cos(h)si...

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What are the stationary points of the curve (1/3)x^3 - 2x^2 + 3x + 2 and what is the nature of each stationary point.

Firstly, a stationary point is a point on the curve where the gradient of the tangent line is equal to 0. Here, this occurs at points where the first derivative of f(x) = 1/3x

Answered by Omar M. • Maths tutor

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Differentiate this equation: xy^2 = sin(3x) + y/x
y² + 2xy dy/dx = 3cos(3x) + 1/y - x/y² dy/dx thereforedy/dx = (3cos(3x) + 1/y - y²) / (2xy + x/y²)

Answered by Alexander E. • Maths tutor
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Can you please explain how to expand two brackets, eg. (3x-7)(5x+6)
When I am expanding brackets I use the acronym FOIL and simplify. You begin by multiply the first term (F) in each bracket, followed by multiplying the outside terms (O), then the inside terms (I) and fin...
Answered by • Maths tutor
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A particle, P, moves along the x-axis. The displacement, x metres, of P is given by 0.5t^2(t^2 - 2t + 1), when is P instantaneously at rest

f(x) = sinx. Using differentiation from first principles find the exact value of f' (π/6).

What are the stationary points of the curve (1/3)x^3 - 2x^2 + 3x + 2 and what is the nature of each stationary point.

Differentiate this equation: xy^2 = sin(3x) + y/x

Can you please explain how to expand two brackets, eg. (3x-7)(5x+6)

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