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Can you check my differentiation?

y=ln 0.2x

y=ln z and z=0.2x

(dy/dx)=1/z

(dz/dy)=0.2

(dy/dx)=(1/(0.2x))*0.2

=0.2/(0.2x)

=1/x

Cheers
Reply 1
Avatar for Zacken
Zacken
22
Original post by Brainfrozen
y=ln 0.2x

y=ln z and z=0.2x

(dy/dx)=1/z

(dz/dy)=0.2

(dy/dx)=(1/(0.2x))*0.2

=0.2/(0.2x)

=1/x

Cheers


Yeah, that's correct. You can see it is because ln 0.2x = ln 0.2 + ln x = constant + ln x

So the derivative of that is the same as the derivative of ln x.
Reply 2
Avatar for Brainfrozen
Brainfrozen
OP
1
Original post by Zacken
Yeah, that's correct. You can see it is because ln 0.2x = ln 0.2 + ln x = constant + ln x

So the derivative of that is the same as the derivative of ln x.


Cheers!
Reply 3
Avatar for Zacken
Zacken
22
Original post by Brainfrozen
Cheers!


No problem.
Reply 4
Avatar for Brainfrozen
Brainfrozen
OP
1
Now I have another problem and my assigned lecturer basically says he's too busy to help me, even although this module of the course is costing 500 pounds. Hardly had any contact with him and he says he's too busy marking assessments. Unreal.

I have gotten as far as this;

(2x + 1)^3 x 6x + (3x^2 - 1) x 6(2x + 1)^2

It then says extract common factors, which I thought I knew what that meant but I'm confused by their result;

6(2x + 1)^2 ((2x + 1)x + (3x^2 - 1))

I'm not a stupid person, but with no support from my lecturer I feel stupid and depressed and with a huge possibility now of not getting pass the first module of this HNC - and I already have an HNC in the bag.
Original post by Brainfrozen
Now I have another problem and my assigned lecturer basically says he's too busy to help me, even although this module of the course is costing 500 pounds. Hardly had any contact with him and he says he's too busy marking assessments. Unreal.

I have gotten as far as this;

(2x + 1)^3 x 6x + (3x^2 - 1) x 6(2x + 1)^2

It then says extract common factors, which I thought I knew what that meant but I'm confused by their result;

6(2x + 1)^2 ((2x + 1)x + (3x^2 - 1))

I'm not a stupid person, but with no support from my lecturer I feel stupid and depressed and with a huge possibility now of not getting pass the first module of this HNC - and I already have an HNC in the bag.


6x(2x+1)3+6(2x+1)2(3x2−1)6x(2x+1)^3+6(2x+1)^2(3x^2-1)

Let a=2x+1⇒6xa3+6a2(3x2−1)a=2x+1 \Rightarrow 6xa^3+6a^2(3x^2-1)

Factor the 6a26a^2

6a2(ax+(3x2−1))6a^2(ax+(3x^2-1))

Back substitute
Reply 6
Avatar for Brainfrozen
Brainfrozen
OP
1
Original post by RDKGames
6x(2x+1)3+6(2x+1)2(3x2−1)6x(2x+1)^3+6(2x+1)^2(3x^2-1)

Let a=2x+1⇒6xa3+6a2(3x2−1)a=2x+1 \Rightarrow 6xa^3+6a^2(3x^2-1)

Factor the 6a26a^2

6a2(ax+(3x2−1))6a^2(ax+(3x^2-1))

Back substitute


Thanks again mate, easy to understand when someone explains it properly. Why my lecturer couldn't spend two minutes doing that - I have no idea.
Reply 7
Avatar for Brainfrozen
Brainfrozen
OP
1
I'm fine with converting between very simple primitives and derivatives, but can someone help me with this?

change this primitive to function: -1/2x^2
change this function to primitive: 1/x^2

baby steps will be much appreciated.

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