How to differentiate?

fatron

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Jan 13, 2020
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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
 

MarkFL

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I would implicitly differentiate with respect to \(x\), and write the equation as:

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

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?
 

Subhotosh Khan

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

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....
 

Dr.Peterson

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Please show what you get when you solve for y in terms of x -- there are a couple tricks you may not know, and we don't yet know where you need help. If you manage that, show your work for the derivative similarly. Just show as much work as you can do, and that will tell us what help you need.
 

Jomo

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I agree with MarkFL thinking that one does not need to solve for y. The problem might even be in the implicit differentiation section.
 

MarkFL

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I agree with MarkFL thinking that one does not need to solve for y. The problem might even be in the implicit differentiation section.
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:

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

Since:

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

We obtain:

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

fatron

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Jan 13, 2020
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Thank you very much everyone, I'm just glancing at this all now but I will sit down and go through it all ASAP and will post if I have any problems. Thank you all.
 
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