How do you differentiate f(x)=xarctansqrtx? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer NJ Apr 4, 2018 =>arctan(sqrtx) + sqrt(x)/(2(1+x)) Explanation: f(x) = x arctan (sqrtx) f'(x) = (d(x))/(dx) arctan(sqrtx) + x (d(arctan(sqrtx)))/(dx) = arctan(sqrtx) + x(1/(1+(sqrtx)^2))(d(sqrtx))/(dx) = arctan(sqrtx) + x/(1+x)*1/(2sqrtx) = arctan(sqrtx) + sqrt(x)/(2(1+x)) Answer link Related questions What is the derivative of f(x)=sin^-1(x) ? What is the derivative of f(x)=cos^-1(x) ? What is the derivative of f(x)=tan^-1(x) ? What is the derivative of f(x)=sec^-1(x) ? What is the derivative of f(x)=csc^-1(x) ? What is the derivative of f(x)=cot^-1(x) ? What is the derivative of f(x)=(cos^-1(x))/x ? What is the derivative of f(x)=tan^-1(e^x) ? What is the derivative of f(x)=cos^-1(x^3) ? What is the derivative of f(x)=ln(sin^-1(x)) ? See all questions in Differentiating Inverse Trigonometric Functions Impact of this question 1506 views around the world You can reuse this answer Creative Commons License