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Find the equation of the tangent of the curve y = (8x)/(x-8) at the point (0,0)

We will be using the quotient rule, although the product rule is also usable and can be run through if the student wishes. Firstly, define u = 8x, v = x-8 for simplicity. Then clearly u' = 8, v' = 1, and ...

TK
Answered by Timofey K. Maths tutor
3415 Views

Differentiate sin(x^3) with respect to y

For this we must use the chain rule. We start by defining x3 as a new variable, u = x3 Can then rewrite the expression as y = sin(u) Chain rule tells us that dy/dx = (dy/du)(du/dx) W...

LB
Answered by Lloyd B. Maths tutor
6443 Views

Solve the simultaneous equations; 3x+2y=11 and 2x-2y=14

2y=11-3x    and 2y=2x-14

11-3x=2x-14

25=5x   thus x=5

3(5) +2y=11

2y=-4    thus y=-2

HS
Answered by Harrie S. Maths tutor
6631 Views

Expand the following expression: -5x(x-7)(x+3)

Expansion of the given expression will result in x to the 3rd power, as there are 3 parts involving x, multiplied together. Let's ignore -5x for a second, and multiply the two expressions in brackets (usi...

MS
Answered by Matas S. Maths tutor
4767 Views

The curve has equation y = x^3 - x^2 - 5x + 7 and the straight line has equation y = x + 7. One point of intersection, B, has coordinates (0, 7). Find the other two points of intersection, A and C.

As both equations are equal to y, we can combine them to create a single equation in terms of x: x^3 - x^2 -5X + 7 = x + 7. Shift the equation so the left hand side is equal to 0 on the right: x^3 - x^2 -...

AB
Answered by Alfie B. Maths tutor
7317 Views

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