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A curve has equation (x+y)^2=x*y^2, find the gradient of the curve at a point where x=1

  1. Differentiating left hand side: 2(x+y)(1+dy/dx) from the chain rule 2. Differentiating right hand side: y2+2xy(dy/dx) from the product rule 3. Equating sides and taking out factors of dy/...
PK
Answered by Peter K. Maths tutor
4421 Views

Can you solve 18-7x < 12-3x ?

In order to solve this formula, you would need to collect similar terms to the same side and then simplify to work out the value of 'x'. We can do this by collecting all the 'x' terms to one side first, t...

BF
Answered by Blessing F. Maths tutor
3920 Views

Find the equation of a straight line that passes through the coordinates (12,-10) and (5,4). Leaving your answer in the form y = mx + c

Finding the gradient (m): The gradient is the change in y-axis over the change in x-axis Δy = -10-4= -14        Δx = 12-5=7 Δy/Δx = -14/7 = -2 Ac...

MM
Answered by Martin M. Maths tutor
8918 Views

The polynomial f(x) is define by f(x) = 3x^3 + 2x^2 - 8x + 4. Evaluate f(2).

f(x) = 3x^3 + 2x^2 - 8x + 4

f(2) = 3(2)^3 + 2(2)^2 - 8(2) + 4

f(2) = 3(8) + 2(4) - 8(2) + 4

f(2) = 24 + 8 - 16 +4

f(2) = 20

SW
Answered by Sophie W. Maths tutor
4088 Views

If y = 2/3 x^3 + x^2; a) What is dy/dx? b) Where are the turning points? c) What are the nature of the turning points?

a) By simple parametric differentiation of each term, dy/dx = 2x^2 + 2x b) The condition for a turning point is the gradient (dy/dx) at that point is zero. 0 = x(2x + 2) so either x=0 or 2x+2=0. Therefore...

AF
Answered by Annabel F. Maths tutor
4421 Views

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