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# Quick Differentiation Q

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1. Hi guys. Mind seems to have gone blank on me. Can anyone help me differentiate the following;

T=16t - t[(1-t)^(-2)]

Just a bit stuck on how to differentiate the (1-t) part. Any help welcome.

Also. Apologies for not Latexing - can't seem to get it working.

Thanks
2. (Original post by PrinceUpsb)
Hi guys. Mind seems to have gone blank on me. Can anyone help me differentiate the following;

T=16t - t[(1-t)^(-2)]

Just a bit stuck on how to differentiate the (1-t) part. Any help welcome.

Also. Apologies for not Latexing - can't seem to get it working.

Thanks
To differentiate the -t[(1-t)^(-2)], you first differentiate the t in the front, and then focus on the (1-t)^(-2). You need to use the chain rule. Do you know how to differentiate (1-t)^(-2)?
3. (Original post by PrinceUpsb)
Hi guys. Mind seems to have gone blank on me. Can anyone help me differentiate the following;T=16t - t[(1-t)^(-2)]Just a bit stuck on how to differentiate the (1-t) part. Any help welcome.Also. Apologies for not Latexing - can't seem to get it working.Thanks

I used quotient rule.

ignore that dT/dt, it should be the next line down.
Also, it probably can be simplified further.
After simplification: (1+t)/(1-t)3
4. Quotient or product rule will do.
5. (Original post by rayquaza17)
To differentiate the -t[(1-t)^(-2)], you first differentiate the t in the front, and then focus on the (1-t)^(-2). You need to use the chain rule. Do you know how to differentiate (1-t)^(-2)?
Yeah. (1-t)^(-2) will diff. to 2(1-t)^-3

So my final answer I am getting is;

dT/dt = 16 - 2(1-t)^-3

can you confirm? I am just multiplying that part by the diff of 'the t in front'.

Cheers
6. (Original post by PrinceUpsb)
Yeah. (1-t)^(-2) will diff. to 2(1-t)^-3

So my final answer I am getting is;

dT/dt = 16 - 2(1-t)^-3

can you confirm? I am just multiplying that part by the diff of 'the t in front'.

Cheers
Using quotient rule is better here.
And no it is not true that (fg)'=f'g' as it seems you have done.
If you want to use product rule you have to use (fg)'=gf'+fg'.

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