Question

How do you find the derivative of the function `f(x)=x+sqrtx`?

Answer 1

Derivative of the function, `f(x)=x+sqrtx,` is `1+1/(2sqrtx).`

Explanation:

Let, `y=f(x)=x+sqrtx.`

`:.`Differentiating `y` w.r.t `x` is,
`d/(dx)(y)=dy/dx=d/(dx)(x+sqrtx)=d/(dx)(x+x^(1/2)).`
`:.dy/dx=1+1/2*x^(1/2-1)=1+1/2x^(-1/2).`
`:.dy/dx=1+(1/2)1/x^(1/2)=1+1/(2sqrtx).`

Therefore, Derivative of the function, `f(x)=x+sqrtx,` is `1+1/(2sqrtx).` (answer).

Answer 2

`dy/dx=f'(x)=1+sqrt(x)/(2x)`

Explanation:

`color(blue)("Given: "f(x)=x+sqrt(x))`

Being of the old school I will use the Leibnitz notation

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Note the general rule of `d/(dx)(x^n) = nx^(n-1)`

set`" " y=x+sqrt(x)`

Then `" "dy/dx=d/(dx)(x)+d/dx(sqrt(x))`

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`color(blue)("Consider ")d/dx(x) -> d/dx(x^1) = 1xx x^0 = 1`

.......................................................................................................
`color(blue)("Consider ")d/dx(sqrt(x))->d/dx(x^(1/2)) = 1/2x^(-1/2) = 1/(2sqrt(x))`

I do not like roots in the denominator so lets see if we can get rid of it.

Multiply by 1 but in the form of `1=sqrt(x)/sqrt(x)` giving:

`1/(2sqrt(x))xxsqrt(x)/sqrt(x) = sqrt(x)/(2x)`
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`color(blue)("Putting it all together")`

`dy/dx=f'(x)=1+sqrt(x)/(2x)" "` which is the same as `1+1/(2sqrt(x))`