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Find the coordinate of the stationary point on the curve y = 2x^2 + 4x - 5.

The important point in the question is the term 'stationary point'. This is where the graph of y will 'flattern out'. If we look at this graph, we can say that the gradient is equal to 0 at this point. Th...

Answered by Serkan M. • Maths tutor

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Using the identity cos(A+B)= cosAcosB-sinAsinB, prove that cos2A=1-2sin^2A.

Use cos(A+B)=cosAcosB-sinAsinB and let A=B so cos(A+A)=cosAcosA-sinAsinA this means cos(2A)=cos²A-sin²A and since cos²A+sin²A=1, cos²A=1-sin²

Answered by Rebecca F. • Maths tutor

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A curve has the equation: x^3 - x - y^3 - 20 = 0. Find dy/dx in terms of x and y.

x³ - x - y³ - 20 = 0 Find dy/dx. Differentiate with respect to x.
3x² - 1 - 3y²(dy/dx) = 0Therefore: dy/dx = (3x

Answered by Karishma P. • Maths tutor

2980 Views

Find the exact gradient of the curve y = ln(1-cos 2x) at the point with x-coordinate π/6.

The gradient to the curve is given by the derivative of the function y = f(x) = ln(1 - cos 2x). This function is a composition of two other functions so we need to use the chain rule to find the derivativ...

Answered by Archie R. • Maths tutor

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