Question

How do you find the derivative of `g(t) = t sqrt(4-t)`?

Answer 1

`g'(t)=(8-3t)/(2sqrt(4-t))`

Explanation:

`"differentiate using "color(blue)"product/chain rules"`

`"Given "g(t)=f(t)h(t)" then"`

`g'(t)=f(t)h'(t)+h(t)f'(t)larrcolor(blue)"product rule"`

`"Given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`f(t)=trArrf'(t)=1`

`h(t)=(4-t)^(1/2)`

`rArrh'(t)=1/2(4-t)^(-1/2)(-1)=-1/2(4-t)^(-1/2)`

`rArrg'(t)=-1/2t(4-t)^(-1/2)+(4-t)^(1/2)`

`color(white)(rArrg'(t))=(4-t)^(-1/2)(-1/2t+4-t)`

`color(white)(rArrg'(t))=(4-t)^(-1/2)(4-3/2t)`

`color(white)(rArrg'(t))=(8-3t)/(2sqrt(4-t))`