dx/dy and dy/dx confusion!! Watch

Funky_Giraffe
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Hey I wonder if you can help - I'm having trouble with this question...

 find\ \frac{dy}{dx} \at\ the\ point\ (2,1)\ on\ the\ graph:


\ x = \sqrt {y^2+3y}

I realise that the rule that dy/dx is the reciprocal of dx/dy, but I am only managing to get dx/dy in terms of y's - therefore I can't just flip my final fraction to get dy/dx since it'll all be in terms of y!! My final answer for dx/dy (in terms of y) was:

 \frac{dx}{dy} = \frac{(2y+3)}{2 \sqrt {y(y+3)}}

But how to get into dy/dx!!!???

Thanks
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09sstinson
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(Original post by Funky_Giraffe)
Hey I wonder if you can help - I'm having trouble with this question...

 find\ \frac{dy}{dx} \at\ the\ point\ (2,1)\ on\ the\ graph:


\ x = \sqrt {y^2+3y}

I realise that the rule that dy/dx is the reciprocal of dx/dy, but I am only managing to get dx/dy in terms of y's - therefore I can't just flip my final fraction to get dy/dx since it'll all be in terms of y!! My final answer for dx/dy (in terms of y) was:

 \frac{dx}{dy} = \frac{(2y+3)}{2 \sqrt {y(y+3)}}

But how to get into dy/dx!!!???

Thanks
just use the value of y=1 you don't need dy/dx in terms of x
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Andy98
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(Original post by Funky_Giraffe)
Hey I wonder if you can help - I'm having trouble with this question...

 find\ \frac{dy}{dx} \at\ the\ point\ (2,1)\ on\ the\ graph:


\ x = \sqrt {y^2+3y}

I realise that the rule that dy/dx is the reciprocal of dx/dy, but I am only managing to get dx/dy in terms of y's - therefore I can't just flip my final fraction to get dy/dx since it'll all be in terms of y!! My final answer for dx/dy (in terms of y) was:

 \frac{dx}{dy} = \frac{(2y+3)}{2 \sqrt {y(y+3)}}

But how to get into dy/dx!!!???

Thanks
If it was me I'd rearrange the original equation
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Funky_Giraffe
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(Original post by 09sstinson)
just use the value of y=1 you don't need dy/dx in terms of x
Of course! I'm so dumb lol.... so do I just plug in 1 to the final dx/dy I got below? And then that will tell me the gradient at that point?
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09sstinson
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(Original post by Funky_Giraffe)
Of course! I'm so dumb lol.... so do I just plug in 1 to the final dx/dy I got below? And then that will tell me the gradient at that point?
just filp the fraction that you got for dx/dy to get dy/dx in terms of y. then put in y=1
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Funky_Giraffe
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(Original post by 09sstinson)
just filp the fraction that you got for dx/dy to get dy/dx in terms of y. then put in y=1
Great! Thanks very much all!
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09sstinson
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(Original post by Funky_Giraffe)
Great! Thanks very much all!
no problem bro
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