At least I don't have to learn anything new!

Regards,

dadon

]]>If *k* is a function with rule of the form k(x) = g(f(x)), where *f* and *g* are smooth functions, then *k* is smooth and

k'(x) = g'(f(x)).f'(x)

(or using Leibniz form: If y = g(u), where u = f(x), then dy/dx = (dy/du)(du/dx) ) .

So, for instance, if you have k(x) = ln(x^2 + 1), then y = g(u), where g(u) = lnu and u = x^2 + 1. This gives k'(x) = (1/(x^2 + 1)).2x = 2x/(x^2 + 1) .

Hope that helps.

]]>Could you explain the composite rule as I havn't come across that.

Thanks agian!:)

]]>1) d/dt(4/t +4lnt) = -4/t^2 + 4/t

2) d/dt(sin(t^2 + 1)) = 2cos(t^2 + 1)t

d/dt(cos(2t - 3)) = -2sin(2t - 3) (both using the composite rule)

3) d/dt(2(e^(-t/2))cos(2t)) = (-e^(-t/2))cos(2t) - 4(e^(-t/2))sin(2t) (using the product rule)

4) I'm not sure about this one, it's a little bit unclear what you mean, but you get the general idea so far. Hopefully the rest are okay, but it's late so i'd check em!

Oh math, i've just realised you've specified t=1.2, so you'd have to use that value with the derived functions to get your answers.

]]>when t = 1.2 for all the following questions:

1) y = 4/t + 4lnt

2) x = sin(t^2 +1) , y = cos(2t -3)

3) x = 1 / 1+2t , y = t / 1+t

4) q = 2e^-t/2.cos2t (e is exponential for this question)

5) x = e^2t.t^3(2-t)^4 (e is exponential for this question)

If someone could help I would be very greatful!

Thanks

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