- Thread starter fatron
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\(\displaystyle x=18(113-200y)^{-1}\)

Thus:

\(\displaystyle 1=-18(113-200y)^{-2}\left(-200\d{y}{x}\right)\)

And so:

\(\displaystyle \d{y}{x}=\frac{(113-200y)^{2}}{3600}\)

Equating this to 1, we get:

\(\displaystyle \frac{(113-200y)^{2}}{3600}=1\)

Can you proceed?

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Assuming 'y' and 'x' as the axes of nominal orthogonal co-ordinate system - gradient would be defined to be 'dy/dx'Sorry everyone,

Can't do this one, just trying find the coordinates where the gradient would equal 1, equation is x=18/(113-200y)

Help with the answer and the method would be great.

Thanks

If I were to do this problem- I would first convert the given function to y = f(x) and then calculate dy/dx and continue....

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Initially I was going to suggest differentiating with respect to \(y\), since the value required for the gradient (1) is the multiplicative inverse of itself, but I thought this very well may be a problem where implicit differentiation is expected. But had I gone with my initial approach, we'd have:I agree with MarkFL thinking that one does not need to solve for y. The problem might even be in the implicit differentiation section.

\(\displaystyle \d{x}{y}=\frac{3600}{(113-200y)^2}=1\)

Since:

\(\displaystyle \d{y}{x}=\frac{1}{\d{x}{y}}\)

We obtain:

\(\displaystyle \left(\frac{113-200y}{60}\right)^2=1\)