lizzpalmer said:
g(t) = (t^3 + t - 2) (2t - 1) ^-5
g ' (t) = (3t + 1) 2 * 5 (2t - 1)^-6 + (t^3 + t - 2) * (-10t^-6)
=(30t + 10)(2t-1)^-6 + (t^3 + t - 2) * 60t^-7
Would that be right? Or what am I doing wrong?
When you are starting out, it is much easier to use substitutions although it is longer. Once this is all familiar, you can skip the substitutions.
Let u = t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + t - 2.
du/dt = 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 1. Polynomial rule.
Let v = 2t - 1
dv/dt = 2. Polynomial rule.
Let w = v[sup:8b0f5wmp]-5[/sup:8b0f5wmp] = (2t - 1)[sup:8b0f5wmp]-5[/sup:8b0f5wmp].
dw/dv = -5v[sup:8b0f5wmp]-6[/sup:8b0f5wmp] = -5(2t - 1)[sup:8b0f5wmp]-6[/sup:8b0f5wmp]. Polynomial rule.
dw/dt = (dw/dv) * (dv/dt) = -10(2t - 1)[sup:8b0f5wmp]-6[/sup:8b0f5wmp]. Chain rule.
Let F(t) = x = uw.
F'(t) = dx/dt = u(dw/dt) + w(du/dt). Product rule.
F'(t) = (t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + t - 2)(-10(2t - 1)[sup:8b0f5wmp]-6[/sup:8b0f5wmp]) + (2t - 1)[sup:8b0f5wmp]-5[/sup:8b0f5wmp](3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 1) =
(-10(t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + t - 2) + (2t - 1)(3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 1)) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp] =
(-10t[sup:8b0f5wmp]3[/sup:8b0f5wmp] - 10t + 20 + 6t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + 2t - 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] - 1) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp]=
(-4t[sup:8b0f5wmp]3[/sup:8b0f5wmp] - 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] - 8t + 19) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp] =
-(4t[sup:8b0f5wmp]3[/sup:8b0f5wmp] + 3t[sup:8b0f5wmp]2[/sup:8b0f5wmp] + 8t - 19) / (2t - 1)[sup:8b0f5wmp]6[/sup:8b0f5wmp].
The least error-prone way to find the derivative of a complex expression is to use substitutions and then use the basic rules of differentiation such as the product rule, the quotient rule, and the chain rule. As I said above, with more experience, you can skip the substitutions.