lets say we need to find a 2nd derivative and we have a following equation
[imath]y^4-2x=5[/imath]
at point (-2, 1)
so we find the first derivative
[math]4y^3\cfrac{dy}{dx}-2=0[/math][math]4y^3\cfrac{dy}{dx}=2[/math]
so our first derivative is [imath]\cfrac{1}{2y^3}[/imath]
now to the second:
[math]\cfrac{-6y^2\cfrac{1}{2y^3}}{4y^6}[/math]
now here is where I find it a bit confusing. why do we insert the point values (only y in this case, so 1) here and not later?
[math]\cfrac{-6*\cfrac{1}{2}}{4} = -\cfrac{3}{4}[/math]why we don't continue with the following route
[math]\cfrac{-\cfrac{6y^2}{2y^3}}{4y^6}=\cfrac{-\cfrac{3}{y}}{4y^6}[/math][math]\cfrac{-\cfrac{6y^2}{2y^3}}{4y^6}=\cfrac{-\cfrac{3}{y}}{4y^6}=-\cfrac{3}{4y^7}[/math]fitting 1 in and getting the same [imath]-\cfrac{3}{4}[/imath]? Is it always the case and going the first route just saves time?
[imath]y^4-2x=5[/imath]
at point (-2, 1)
so we find the first derivative
[math]4y^3\cfrac{dy}{dx}-2=0[/math][math]4y^3\cfrac{dy}{dx}=2[/math]
so our first derivative is [imath]\cfrac{1}{2y^3}[/imath]
now to the second:
[math]\cfrac{-6y^2\cfrac{1}{2y^3}}{4y^6}[/math]
now here is where I find it a bit confusing. why do we insert the point values (only y in this case, so 1) here and not later?
[math]\cfrac{-6*\cfrac{1}{2}}{4} = -\cfrac{3}{4}[/math]why we don't continue with the following route
[math]\cfrac{-\cfrac{6y^2}{2y^3}}{4y^6}=\cfrac{-\cfrac{3}{y}}{4y^6}[/math][math]\cfrac{-\cfrac{6y^2}{2y^3}}{4y^6}=\cfrac{-\cfrac{3}{y}}{4y^6}=-\cfrac{3}{4y^7}[/math]fitting 1 in and getting the same [imath]-\cfrac{3}{4}[/imath]? Is it always the case and going the first route just saves time?
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