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# Derivatives watch

1. Hi all,
This is the only part of my maths course i dont fully understand now.

Ive got two questions where i need to find the first derivative.

can someone give me solutions to

1) y = 4e^2x^2 - 3x + 1
2) y = loge(x^2 - 6x + 9)

Can you explain how you got to the solutions too, i want to use the solutions to these to help me find the first derivitive to other questions of a similar nature.

Also is there a way you can check the answers?

cheers
gav
2. (Original post by gavo)
Hi all,
This is the only part of my maths course i dont fully understand now.

Ive got two questions where i need to find the first derivative.

can someone give me solutions to

1) y = 4e^2x^2 - 3x + 1
2) y = loge(x^2 - 6x + 9)

Can you explain how you got to the solutions too, i want to use the solutions to these to help me find the first derivitive to other questions of a similar nature.

Also is there a way you can check the answers?

cheers
gav
To solve 1) and 2) use substitution and the chain rule.

1) y = 4e^(2x²) - 3x + 1

for the e^(2x²) bit, let v = e^(2x²), let u = 2x² , so v = e^u

then
dv/du = e^u
du/dx = 4x

so,
dv/dx = dv/du.du/dx (chain rule)
dv/dx = e^u.(4x)
dv/dx = 4xe^(2x²)
==============

y = 4v - 3x + 1 (using earlier substitution)
dy/dx = 4.dv/dx - 3 + 0
dy/dx = 16xe^(2x²) - 3
==================

2) y = loge(x^2 - 6x + 9)

let u = x^2 - 6x + 9
du/dx = 2x - 6
===========

y = loge(u)
dy/dx = dy/du.du/dx (chain rule)
dy/dx = (1/u).(2x-6)
dy/dx = (2x-6)/(x^2 - 6x + 9)
=======================
3. (Original post by gavo)

1) y = 4e^2x^2 - 3x + 1
2) y = loge(x^2 - 6x + 9)

cheers
gav
For 2) note that y = ln((x-3)²) = 2ln(x-3) => dy/dx = 2/(x-3)

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