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

Question

Paisley Johnson · Accepted Answer

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

Christopher Allers · Answer

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

#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))#

Emily Ammerman · Answer

undefined To find the derivative of the function $ f(x) = x + \sqrt{x} $:

1. Differentiate each term separately.
2. For $ x $, the derivative is $ 1 $ since it follows the power rule ($ \frac{d}{dx} x^n = nx^{n-1} $).
3. For $ \sqrt{x} $, the derivative is $ \frac{1}{2\sqrt{x}} $ using the power rule and the chain rule.
4. Combine the derivatives of both terms.
5. The derivative of $ f(x) = x + \sqrt{x} $ is $ f'(x) = 1 + \frac{1}{2\sqrt{x}} $.