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P- (3,0) y=ln(x/3)     u=x/3    y=ln(u) ​​​​​​            du = 1/3  dy = 1/u = 3            dx       du dy= du x dy dx dx  du   = 1/3 x 3 = 1 gradient at normal = -1 equ...

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x=3,y=2

Simultaneous equations have two unknowns, you need to find out one of the unkowns (eg. x) in terms of the other (e.g. y) in order to solve both unknowns. Using these two examples (x and y) we can construc...

this is implicit differentiation. We start by differentiating 3x^(2) to get 6x (lower the power by 1, multiply by the original power). For xy, we use the product rule, giving us y + (x)dy/dx (this is the ...