Exponential Question (1 Viewer)

[sasquatch]

Just a question: Find the equation of the tangent to the curve y = e^x + 1 at the point (1, e + 1)

This is my working...

y = e^x + 1

dy/dx = e^x

At (1, e + 1)

dy/dx = e¹
= e

y - (e + 1) = e(x - 1)
y - e - 1 = ex - e
y - 1 = ex
y = ex + 1

well the answer says..

y = e^x + 1

I cant see what i did wrong.. i dont think i did anything wrong, so im guessing its a type since both look the same, except in mine the x is not a power. Could somebody verify this for me. Thanks.

 

[richz]

looks fine to me , the answers are rong

 

[pLuvia]

Just a question: Find the equation of the tangent to the curve y = e^x + 1 at the point (1, e + 1)

This is my working...

y = e^x + 1

dy/dx = e^x

At (1, e + 1)

dy/dx = e¹
= e

y - (e + 1) = e(x - 1)
y - e - 1 = ex - e
y - 1 = ex
y = ex + 1

well the answer says..

y = e^x + 1

I cant see what i did wrong.. i dont think i did anything wrong, so im guessing its a type since both look the same, except in mine the x is not a power. Could somebody verify this for me. Thanks.

That looks right to me, answer is wrong. Probably typo

 

[SoulSearcher]

Just a question: Find the equation of the tangent to the curve y = e^x + 1 at the point (1, e + 1)

This is my working...

y = e^x + 1

dy/dx = e^x

At (1, e + 1)

dy/dx = e¹
= e

y - (e + 1) = e(x - 1)
y - e - 1 = ex - e
y - 1 = ex
y = ex + 1

well the answer says..

y = e^x + 1

I cant see what i did wrong.. i dont think i did anything wrong, so im guessing its a type since both look the same, except in mine the x is not a power. Could somebody verify this for me. Thanks.

you can tell its a typo, the equation of the tangent to the curve at a particular point cannot be the equation of the curve itself

 

[sasquatch]

you can tell its a typo, the equation of the tangent to the curve at a particular point cannot be the equation of the curve itself

Yeah it can..

Find the equation of the tangent to y = x at (1,1)

dy/dx = 1

At (1,1)

dy/dx = 1

y - 1 = 1(x - 1)
y - 1 = x - 1
y = x

oh wait a second.. you said curve... so much for me being a smart ass ( i was just kidding anyway)

 

[Riviet]

It's always good to be sure you're sure of something.
Your answer to the first problem is correct.

 

[nallask8r]

yeh got a couple of Q's as well, cheers
Find first derivative of

e^x + 1/e^-x - 1

and

f(x)=(1 + 2x)e^3x, find f'(2) and f''(0)

 

[pLuvia]

y=e^x+1/(e^-x-1)

dy/dx = {e^x + [0 - (-e^-x]} / (e^-x-1)²
= [e^x+e^-x] / (e^-x-1)²

f(x)=(1+2x)e^3x

f'(x)=2e^3x+(3+6x)e^3x
f''(x)=6e^3x+6e^3x+(9+18x)e^3x
= 12e^3x+(9+18x)e^3x

f'(2) = 2e⁶+15e⁶=17e⁶
f''(0) = 12 + 9 = 21

 

[SoulSearcher]

well ok then, I'm a bit rusty but I'll help

using the quotient rule,

d/dx ( e^x + 1 / e^-x - 1 ) = { e^x ( e^-x - 1 ) + e^-x ( e^x + 1 ) } / ( e^-x - 1 )²

if possible, simplify but I'm too lazy to bother

second question

f(x) = (1 + 2x) * e^3x , using chain rule
f'(x) = 2e^3x + 3e^3x(1 + 2x)
= e^3x(5 + 6x)
f''(x) = 3e^3x(5+6x) + 6e^3x
= 3e^3x(7 + 6x)

therefore
f'(2) = e⁶(5 + 12)
f'(2) = 6858.29 to 2 dec. pl.

and
f''(0) = 3e⁰(7 + 0)
f''(0) = 3 * 7
f''(0) = 21

hmm i simplified the second question before subbing in my values.

 

[nallask8r]

thanks heaps..just started the topic for 2unit trying to my head around it

got a couple more this time with integration... i normally just recipracol the derivative but cant get these ones

xe^x^2 dx between 2 and 1

and Find the area bounded by the curve y= e^x, the x and y axes, and the line x=2

 

[SoulSearcher]

thanks heaps..just started the topic for 2unit trying to my head around it

You'll soon find that this exponential stuff gets easier as you do more of it, as with most maths topics

 

[pLuvia]

2
∫ xe^x2 dx
1

For this one I think you would have to use integration by parts, I 4unit topic

Find the area bounded by the curve y= e^x, the x and y axes, and the line x=2

2
∫ e^x dx
0

= [e^x]
Sub in the x values

= e²-1

 

[SoulSearcher]

xe^x^2 dx between 2 and 1

wtf? I've never seen that one in a 2 unit textbook ...

 

[Riviet]

xe^x^2 dx between 2 and 1

For this one I think you would have to use integration by parts, I 4unit topic

Is IBP really needed for this?

Just let u=x²
du/dx=2x
du=2x dx
When x=2, u=4 and when x=1, u=1
So the integral then becomes:

4
∫ 1/2 x e^u du
1
The integral should now be pretty straight forward.

 

[pLuvia]

My mistake

 

[pLuvia]

Might as well finish it

2
∫ xe^x2 dx
1

Let u=x²
du=2x dx

2
1/2 ∫ 2xe^x2 dx
1

2
1/2 ∫ e^u du
1

1/2 [ e^u]

Sub in values

1/2 [e² - e]

 

[nallask8r]

wtf? I've never seen that one in a 2 unit textbook ...

lol its in a 3unit mathematics book but we get tested on it