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Quick Differentiation Q

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

Reply 1

rayquaza17

Original post by PrinceUpsb

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)?

Reply 2

jackien1

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

Do you have the answer?

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

(edited 7 years ago)

Reply 3

B_9710

Quotient or product rule will do.

Reply 4

PrinceUpsb

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

Reply 5

IrrationalRoot

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'.