I need to find the derivative of
$\displaystyle x-2 \sqrt{x-1}$
I knew to use the power rule, but I was coming up with
$\displaystyle 1/2(x-1)^{-1/2}$ Is this the correct answer?
Thanks
Jason
I need to find the derivative of
$\displaystyle x-2 \sqrt{x-1}$
I knew to use the power rule, but I was coming up with
$\displaystyle 1/2(x-1)^{-1/2}$ Is this the correct answer?
Thanks
Jason
chain rule is present at any moment.
for example if you want to differentiate a simple function as $\displaystyle x^2,$ then that is $\displaystyle 2x\cdot(x)'=2x,$ but that it's quite simple and then irrelevant to write, but having $\displaystyle \sqrt{x-1}$ we can think this as the composition by letting $\displaystyle f(x)=\sqrt x$ and $\displaystyle g(x)=x-1,$ so by differentiating $\displaystyle f\big(g(x)\big)$ we get $\displaystyle f'\big(g(x)\big)\cdot g'(x).$ According to this we have that $\displaystyle \sqrt{x-1}$ can be expressed as $\displaystyle h(x)=f\big(g(x)\big)=\sqrt{x-1},$ and then by differentiating we get (following up on what I said before) $\displaystyle h'(x)=\frac1{2\sqrt{x-1}}\cdot(x-1)'=\frac1{2\sqrt{x-1}}.$