Question

How do you find the derivative of `sqrt(5-3x)`?

Answer 1

`=(-3)/(2sqrt(5-3x))`

Explanation:

To find the derivative of the expression is by applying chain rule
Let `v(x)=5-3x and u(x)=sqrtx`

Then `u@v(x)=u(v(x))=sqrt(5-3x)`

Then `color(blue)((sqrt(5-3x))'=(u@v(x))'=u'(v(x))xxv'(x)`

Computing `u'(v(x)) and v'(x)`

`u'(x)=1/(2sqrtx)`
Then

`color(blue)( u'(v(x))=1/(2sqrt(5-3x))`

`color(blue)(v'(x)=-3`

`color(blue)((sqrt(5-3x))'=(u@v(x))'=u'(v(x))xxv'(x)`

`(sqrt(5-3x))'=1/(2sqrt(5-3x))xx-3`

Therefore,
`(sqrt(5-3x))'=(-3)/(2sqrt(5-3x))`

Answer 2

`d/dx sqrt(5-3x) = -3/(2sqrt(5-3x)`

Explanation:

`f(x)= sqrt(5-3x)`
We know,
`d/dx(sqrtx) = 1/(2sqrtx)`
and also,
`d/dx(a-bx)=-b`

hence, using the chain rule, we differentiate `f(x)` to get
`d/dx sqrt(5-3x) = -3/(2sqrt(5-3x)`

Answer 3

The answer is `=-3/(2sqrt(5-3x))`

Explanation:

For this, we use `(sqrtu)'=1/(2sqrtu)` and the chain rule

So, `(sqrt(5-3x))'=1/(2sqrt(5-3x))*(-3x)'`

`=-3/(2sqrt(5-3x))`

Answer 4

`-3/(2sqrt(5-3x))`

Explanation:

Express `y=sqrt(5-3x)=(5-3x)^(1/2)`

differentiate using the `color(blue)"chain rule"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(dy)/(du)xx(du)/(dx))color(white)(2/2)|)))to(A)`

let `u=5-3xrArr(du)/(dx)=-3`

and `y=u^(1/2)rArr(dy)/(du)=1/2u^(-1/2)`

substitute these values into (A) changing u back into terms of x.

`rArrdy/dx=1/2u^(-1/2)xx(-3)=-3/(2u^(1/2))=-3/(2sqrt(5-3x))`