Derivative: Application of the Product rule

Question

Suppose we have a function $\pi_1(t|v_1)=b^*(t)^{n-1}(v_1-b^*(t))$

When I differentiate this, I need to use the chain rule and the product ruke,

$(n-1)b^*(t)^{n-2}\frac{db^*}{dt}(v_1-b^*(t)) - b^*(t)^{n-1}\frac{db^*}{dt}$

As you can see I apply the chain rule to the first part of the function and the rest is an application of the product rule? Is this correct?

Answer 1

$\pi_1(t|v_1)=b^*(t)^{n-1}(v_1-b^*(t))$

As $\pi_1(t|v_1) = \color{red}{b^*(t)^{n-1}}\times \color{blue}{(v_1-b^*(t))}$ so according the product rule

$\frac {d\pi_1'(t|v_1)}{dt} = \frac {d\color{red}{b^*(t)^{n-1}}}{dt}[\color{blue}{(v_1-b^*(t))}] + [\color{red}{b^*(t)^{n-1}}]\frac {d\color{blue}{(v_1-b^*(t))}}{dt}$

$\color{red}{b^*(t)^{n-1}} = \color{green}(\color{orange}{b^*(t)}\color{green}{)^{n-1}}$ so according to the chain rule

$\frac {d\color{red}{b^*(t)^{n-1}}}{dt}=\frac {d\color{green}(\color{orange}{b^*(t)}\color{green}{)^{n-1}}}{d\color{orange}{b^*(t)}}\frac {d\color{orange}{b^*(t)}}{dt}=\color{green}{(n-1)(}\color{orange}{b^*(t)}\color{green}{)^{n-2}}\frac {d\color{orange}{b^*(t)}}{dt}$

And $\color{blue}{(v_1-b^*(t))} = \color{purple}{v_1} - \color{orange}{b^*(t)}$ so according to the sum/difference rule

$\frac {d\color{blue}{(v_1-b^*(t))}}{dt} = \frac {d\color{purple}{v_1}}{dt} - \frac {d\color{orange}{b^*(t)}}{dt}= \color{purple}0 - \frac {d\color{orange}{b^*(t)}}{dt}=-\frac {d\color{orange}{b^*(t)}}{dt}$

Putting it all together we get exactly what you got

$\frac {d\pi_1'(t|v_1)}{dt} = \color{green}{(n-1)(}\color{orange}{b^*(t)}\color{green}{)^{n-2}}\frac {d\color{orange}{b^*(t)}}{dt}\color{blue}{(v_1-b^*(t))} - \color{red}{b^*(t)^{n-1}}\frac {d\color{orange}{b^*(t)}}{dt}$

I'm not comfortable with you language of "parts" but I imagine what you meant to say was correct. The above is more detail then you probably need but it's good practice to really run a comb through things to make sure one really understands.

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Oh we can further factor this to get

$\frac {d\pi_1'(t|v_1)}{dt} = [(n-1)b^*(t)^{n-2}(v_1-b^*(t)) - b^*(t)^{n-1}]\frac {db^*(t)}{dt}=[(n-1)(v_1-b^*(t))- b^*(t)]b^*(t)^{n-2}\frac{db^*(t)}{dt}$

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Alternative as

$\pi_1(t|v_1) = b^*(t)^{n-1}(v_1-b^*(t)) = v_1b^*(t)^{n-1} - b^*(t)^n$ we can calculate the derivative as

$\frac {\pi_1(t|v_1) }{dt} = v_1(n-1)b^*(t)^{n-2}\frac{db^*(t)}{dt} - nb^*(t)^{n-1}\frac{db^*(t)}{dt}$