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# Differentiation P3 watch

1. How do you differentiate with respect to x" (x-1)^3 e^(x^2)-2x

I get stuck because of the e^(x^2)-2x bit.
2. (Original post by TheQueen1986)
How do you differentiate with respect to x" (x-1)^3 e^(x^2)-2x

I get stuck because of the e^(x^2)-2x bit.
Do you mean e^(x^2 - 2x)?

If so: y = (x - 1)^3.e^(x^2 - 2x)

u = (x - 1)^3, v = e^(x^2 - 2x)

Applying the product rule: dy/dx = vdu/dx + udv/dx (and chain rule actually, cos you have a function of a function)

dy/dx = [e^(x^2 - 2x)].3(x - 1)^2.1 + [(x - 1)^3].dv/dx

v = e^(x^2 - 2x)

Let t = x^2 - 2x, then v = e^t

By the chain rule, dv/dx = (dv/dt)(dt/dx)

dv/dx = e^t.(2x - 2)

dv/dx = (2x - 2)e^(x^2 - 2x)

=> dy/dx = [e^(x^2 - 2x)].3(x - 1)^2.1 + [(x - 1)^3].(2x - 2)e^(x^2 - 2x)
3. Thanks I get that now. It was similar to the y=ln ... problem.

The answer at the back wants it in this form: (x-1)^2 e(x^2-2x)[2x^2-4x+5]. Does it matter?
4. (Original post by TheQueen1986)
Thanks I get that now. It was similar to the y=ln ... problem.

The answer at the back wants it in this form: (x-1)^2 e(x^2-2x)[2x^2-4x+5]. Does it matter?
Well, you could just put it in that form:

=> dy/dx = [e^(x^2 - 2x)].3(x - 1)^2.1 + [(x - 1)^3].(2x - 2)e^(x^2 - 2x)

Take out common factors (ie. (x - 1)^2 and e^(x^2 - 2x)):

dy/dx = [(x - 1)^2.e^(x^2 - 2x)][3 + (x - 1)(2x - 2)]

= [(x - 1)^2.e^(x^2 - 2x)](3 + 2x^2 - 2x - 2x + 2)

= [(x - 1)^2.e^(x^2 - 2x)](2x^2 - 4x + 5) as required.

Not sure it matters too much, as both are correct.
5. the question is
e^(xy)=4
i tried a lot. is there anyone who can help? :
6. (Original post by bodomx)
the question is
e^xy=4
i tried a lot. is there anyone who can help? :
Is it
(e^x)y = 4 ?
or
e^(xy)=4 ?

If it’s (e^x)y = 4 then …

Divide both sides by e^x, giving
y = 4e^(-x)
dy/dx = 4*(-1)*e^(-x)
dy/dx = -4/(e^x)
=============

If it’s e^(xy) = 4 then …

use substitution and the product rule.

Let u = xy

du/dx = x.dy/dx + y.(1)

Then,

e^u = 4

differentiating both sides wrt x, using the chain rule,

d(e^u)/dx = d(4)/dx
d(e^u)/du * du/dx = 0
e^u * (x.dy/dx + y) = 0
x.dy/dx + y = 0
dy/dx = -y/x
=========
7. I got another one, here it is
Find the value of sinθ when 7secθ-3tanθ is minimum.
I did like this> first I differenciated 7secθ-3tanθ then equated it to zero. I couldn't get the answer this way. Then I used another method. I equated the 7secθ-3tanθ to zero. And then divided both sides by secθ. Got the book answer this way. but i think this makes no sense since the question says when it is at minimum. Anyone got an idea about this?
8. (Original post by bodomx)
I got another one, here it is
Find the value of sinθ when 7secθ-3tanθ is minimum.
I did like this> first I differenciated 7secθ-3tanθ then equated it to zero. I couldn't get the answer this way. Then I used another method. I equated the 7secθ-3tanθ to zero. And then divided both sides by secθ. Got the book answer this way. but i think this makes no sense since the question says when it is at minimum. Anyone got an idea about this?
erm - you need to rearrange the 7sec0-3tan0 (i dont have a theta key ) into a simpler trig function to differentiate it - did you do that?
9. (Original post by bodomx)
I got another one, here it is
Find the value of sinθ when 7secθ-3tanθ is minimum.
I did like this> first I differenciated 7secθ-3tanθ then equated it to zero. I couldn't get the answer this way. Then I used another method. I equated the 7secθ-3tanθ to zero. And then divided both sides by secθ. Got the book answer this way. but i think this makes no sense since the question says when it is at minimum. Anyone got an idea about this?
If you equate 7secθ-3tanθ to zero, then you get,

7secθ-3tanθ = 0
7 - 3sinθ = 0
sinθ = 7/3

but sinθ is never greater than 1! So there's a mistake in doing it this way.

My way!
======

let f = 7secθ-3tanθ
f' = 7secθ.tanθ - 3sec²θ = 0 for a turning point.
7secθ.tanθ - 3sec²θ = 0
7tanθ - 3secθ = 0
7tanθ = 3secθ
7sinθ = 3 - multiplying by cosθ
sinθ = 3/7
========
10. (Original post by Fluffstar)
erm - .... (i dont have a theta key ) ...
erm - you can enter θ into your posts by typing &#952 immediately followed by a semicolon.
If you want any more symbols like that, search these forums for the text "special characters and symbols"

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