How do you determine the values of x at which #sqrt(x^2 + 9)# is differentiable? Calculus Derivatives Differentiable vs. Non-differentiable Functions 1 Answer Konstantinos Michailidis May 29, 2016 The derivative of #f(x)=sqrt(x^2+9)# is #f'(x)=x/(sqrt(x^2+9))# which exists for all real numbers. Answer link Related questions What are non differentiable points for a function? What does differentiable mean for a function? What are non differentiable points for a graph? How do you find the non differentiable points for a function? How do you find the non differentiable points for a graph? What are differentiable points for a function? How do you find the differentiable points for a graph? What is the derivative of a unit vector? On what interval is the function #ln((4x^2)+9)# differentiable? How do you find the partial derivative of the function #f(x,y)=intcos(-7t^2-6t-1)dt#? See all questions in Differentiable vs. Non-differentiable Functions Impact of this question 2875 views around the world You can reuse this answer Creative Commons License