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Inverse (1 Viewer)
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member 6003
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no why do you think that the derivative of inverse f(x) is dx/dy?
)
=x)
=1)
})
)})
 \implies \frac{dy}{dx}=f'(x) )
}{f'(f^{-1}(x))}=-1)
idk how to prove this false
This is kind of useless for this question though. y=-x proves A wrong, B is obviously wrong, y=x proves C wrong,
if you try to draw not the whole graph but just the point of intersection on the line y=x where its perpendicular but without a gradient of 0 and infinity (where derivative doesn't exist) then you can't without not making it asymmetric.
This is kind of useless for this question though. y=-x proves A wrong, B is obviously wrong, y=x proves C wrong,
if you try to draw not the whole graph but just the point of intersection on the line y=x where its perpendicular but without a gradient of 0 and infinity (where derivative doesn't exist) then you can't without not making it asymmetric.
Isnt it because like the gradient of the inverse at (a,b) is the reciprocal of the gradient of the function at (b,a)?no why do you think that the derivative of inverse f(x) is dx/dy?
idk how to prove this false
This is kind of useless for this question though. y=-x proves A wrong, B is obviously wrong, y=x proves C wrong,
if you try to draw not the whole graph but just the point of intersection on the line y=x where its perpendicular but without a gradient of 0 and infinity (where derivative doesn't exist) then you can't without not making it asymmetric.
member 6003
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- Apr 20, 2022
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- 2023
Yeah I think that works