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Help needed, expon and log functions (1 Viewer)

RHINO7

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I can't work out these questions=

i)Find the gradient of the tangent to the curve y=4e^x at the point where x=1.6, correct to 2 decimal places.


ii)Find the equation of the tangent to the curve y=-e^x at the point (1,-e)


iii)Find the first and second derivatives of y= 7e^x. Hence show that the second derivative = y
 

jb_nc

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d(e^x)/dx = e^x

No joke that's all you ned for that question.
 

ellen.louise

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RHINO7 said:
I can't work out these questions=

i)Find the gradient of the tangent to the curve y=4e^x at the point where x=1.6, correct to 2 decimal places.


ii)Find the equation of the tangent to the curve y=-e^x at the point (1,-e)


iii)Find the first and second derivatives of y= 7e^x. Hence show that the second derivative = y
i) y=4e^x
y'=4e^x
m of tan (x=1.6) = 4e^1.6

y-4e^1.6= 4e^1.6(x-1.6)
go from there...

amirite?

iii) y=7e^x
y'=7e^x ??? is there something i am missing here? this seems too easy. i know it works for just e^x, does it work for this too???
 

jb_nc

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they meant for (iii) that d^2y/dx^2 = y = 7e^x which it does. I have a feeling the OP is going to feel pretty stupid for not being able to do these.
 

RHINO7

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yeah thanks, what about the second question
 

ellen.louise

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RHINO7 said:
I can't work out these questions=

i)Find the gradient of the tangent to the curve y=4e^x at the point where x=1.6, correct to 2 decimal places.


ii)Find the equation of the tangent to the curve y=-e^x at the point (1,-e)


iii)Find the first and second derivatives of y= 7e^x. Hence show that the second derivative = y
ii)

y'= -e^x
m of tan = -e

y+e=-e(x-1)
 

cl3nta

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jb_nc is right but here are the solutions anyways :)

i)
dy/dx=4e^x
at x=1.6, dy/dx=4e^1.6 = gradient

ii)
dy/dx=-e^x
at x=1, m=-e
therefore the equation is:
y+e=-e(x-1)
y=-ex

iii)
y=7e^x
dy/dx=7e^x
d^2y/dx^2=7e^x
From above, d^2y/dx^2=y=7e^x QED.

Hope that helps :) wait till you get to integrals ;)
 
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cl3nta

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wow so many posts between when I read the questions to when I posted ><
 

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