Question

Question #c7427

Answer 1

`(2x-3)/(4sqrt(x-3))`

Explanation:

`f(x)=1/2xsqrt(x-3)`
this can be seen as
`1/2x(x-3)^(1/2)`

(Note: it's not possible to solve an equation such as the one you've given. We could try to find its zeros, but since this is placed in calculus `rarr` differentiation rules, this is a guide of how to differentiate the function, or find its derivative.)

we can apply the chain rule and the product rule
first lets apply the product rule. The product rule is defined as
`(fg)prime = fprime*g + f*gprime`

here we can treat
`f =1/2x`
`g =(x-3)^(1/2)`

which leads to
`fprime(x) =1/2(x-3)^(1/2) +1/2x*[ d/(dx) (x-3)^(1/2)]`
you will notice that in brackets we still need to solve the derivative for the last term but this where we apply the chain rule.

`(f(g))prime = fprime(g) * gprime`

this leads to

`fprime(x) =1/2(x-3)^(1/2) +1/2x* 1/2(x-3)^(-1/2)`

`fprime(x) =1/2(x-3)^(1/2) +x/(4(x-3)^(1/2))`

`fprime(x) =1/2(x-3)/(2(x-3)^(1/2)) +x/(4(x-3)^(1/2))`

`fprime(x) =(2x-3)/(4(x-3)^(1/2))`

`fprime(x) =(2x-3)/(4sqrt(x-3))`