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c3 differentiation q

can't seem to figure out q:
Find the equation of the tangent and the normal to the curve y=e^x(lnx) at the point where x=1
I worked the first derivative to be dy/dx=e^x(lnx) + e^x/x
when i sub x=1 to find gradient of tangent i get e but from this point i go wrong !
Original post by Wisteria_xo
can't seem to figure out q:
Find the equation of the tangent and the normal to the curve y=e^x(lnx) at the point where x=1
I worked the first derivative to be dy/dx=e^x(lnx) + e^x/x
when i sub x=1 to find gradient of tangent i get e but from this point i go wrong !


Correct so far.

But how did you go wrong after that? What did you try? :h:
Original post by SeanFM
Correct so far.

But how did you go wrong after that? What did you try? :h:


Gradient of tangent = e
Tried to find the equation of the tangent, x=1 y=0 and c=-2 therefore i got y=ex-e
however the mark scheme from the book said y=ex is eq of tangent.

Finding normal I got gradient = -1/e and c=1/e therefore i got y=-1/ex +1/e
which simplifies to ey=-x+1
again the book gives a different answer ey+x = 1+e^2
Original post by Wisteria_xo
Gradient of tangent = e
Tried to find the equation of the tangent, x=1 y=0 and c=-2 therefore i got y=ex-e
however the mark scheme from the book said y=ex is eq of tangent.

Finding normal I got gradient = -1/e and c=1/e therefore i got y=-1/ex +1/e
which simplifies to ey=-x+1
again the book gives a different answer ey+x = 1+e^2


:hmmmm: is there any chance that f(x)=exlnx f(x) = e^{xlnx} ? You can already see that the equation of the tangent is wrong as f(1) is not 0. It is always possible that the textbook is wrong.
Original post by SeanFM
:hmmmm: is there any chance that f(x)=exlnx f(x) = e^{xlnx} ? You can already see that the equation of the tangent is wrong as f(1) is not 0. It is always possible that the textbook is wrong.


:s-smilie: no the f(x) is written correctly aha i suppose our textbook is famous for its mistakes !
Original post by Wisteria_xo
:s-smilie: no the f(x) is written correctly aha i suppose our textbook is famous for its mistakes !


f(x)=exln⁡(x)⟶f(1)=0⇒f′(x)=exx(xln⁡x+1)⟶f′(1)=ef(x)=e^x\ln(x) \longrightarrow f(1)=0 \Rightarrow f'(x)=\frac{e^x}{x}(x\ln{x} + 1) \longrightarrow f'(1)=e

So tangent is y=e(x−1)y=e(x-1) and normal is y=−1e(x−1)y=-\frac{1}{e}(x-1)

If this is not what the book shows, then it is wrong.
Original post by RDKGames
f(x)=exln⁡(x)⟶f(1)=0⇒f′(x)=exx(xln⁡x+1)⟶f′(1)=ef(x)=e^x\ln(x) \longrightarrow f(1)=0 \Rightarrow f'(x)=\frac{e^x}{x}(x\ln{x} + 1) \longrightarrow f'(1)=e

So tangent is y=e(x−1)y=e(x-1) and normal is y=−1e(x−1)y=-\frac{1}{e}(x-1)

If this is not what the book shows, then it is wrong.


I got exactly the same ! I guess it must be wrong then, thank you so much!

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