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

2 Answers
Nov 2, 2016

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).

Nov 2, 2016

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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))

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Putting it all together")

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